Differentiation Part4

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Parametric Functions

*This part is to be rearranged/redone

Remember that parametric functions will be in the form

Notice that normally, when we have ${\color{Red}y}=f\left({\color{Blue}x}\right)$ , we can find $\frac{d{\color{Red}y}}{d{\color{Blue}x}}$

Imagine we have ${\color{Red}\heartsuit}=f\left({\color{Blue}\bigcirc}\right)$ , we can find $\frac{d{\color{Red}\heartsuit}}{d{\color{Blue}\bigcirc}}$

Thus, since we have $x=f\left(t\right), y=f\left(t \right)$ , we can find

$\frac{dx}{dt}$ AND also
$\frac{dy}{dt}$

But how would be find $\frac{dy}{dx}$ ?

Remember that chain rule give us $\frac{dy}{dx}=\frac{dy}{d{\color{Red}\heartsuit}}\times\frac{d{\color{Red}\heartsuit}}{dx}$

Since we already have $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , thus we will use

$\frac{dy}{dx}=\frac{dy}{{\color{Red}dt}}\times\frac{{\color{Red}dt}}{dx}$

However notice that we have can find $\frac{dx}{dt}$ but we need $\frac{dt}{dx}$

Thus we can use the law either as

$\frac{dy}{dx}=\frac{dy}{dt}\times\frac{1}{\left(\cfrac{dx}{dt}\right)}$
Or directly as $\frac{dy}{dx}=\frac{\left(\cfrac{dy}{dt}\right)}{\left(\cfrac{dx}{dt}\right)}$

However, if we have fraction for $\frac{dx}{dt}$ , it's better we just multiply the reciprocal (rather than divide by a fraction)

if $\frac{dx}{dt}=\frac{1}{a},\; \frac{dt}{dx}$ $=a\,$
if $\frac{dx}{dt}=\frac{a}{b},\; \frac{dt}{dx}$ $=\frac{b}{a}$
if $\frac{dx}{dt}=a+\frac{1}{b}$ $=\frac{ab+1}{b}$ , $\frac{dt}{dx}$ $=\frac{b}{ab+1}$

Note:

Keep in mind you final answer will be in terms $t\,$ or whatever the parameter is
Don't mix up your $x\,$ 's and your $y\,$ 's
If you are not sure which divide by which (for example, during exam), simply write out the chain rule and work it out
Be extra careful (and check) when you have fractions for the $\frac{dy}{dt}$ and/or $\frac{dx}{dt}$

Quick Practice:

$\frac{dx}{dt}=a,\; \frac{dy}{dt}=b,\; \therefore$ $\frac{dy}{dx}=\frac{b}{a}$
$\frac{dx}{dt}=a,\; \frac{dy}{dt}=\frac{1}{b},\; \therefore$ $\frac{dy}{dx}=\frac{1}{ba}$
$\frac{dx}{dt}=\frac{1}{a},\; \frac{dy}{dt}=b,\; \therefore$ $\frac{dy}{dx}=ba$
$\frac{dx}{dt}=\frac{1}{a},\; \frac{dy}{dt}=\frac{1}{b},\; \therefore$ $\frac{dy}{dx}=\frac{a}{b}$
$\frac{dx}{dt}=\frac{a}{b},\; \frac{dy}{dt}=\frac{c}{d},\; \therefore$ $\frac{dy}{dx}=\frac{c}{d}\times \frac{b}{a}=\frac{cb}{ad}$

Beyond that, there isn't really anything new you need to learn here. However, the following examples and exercises WILL serve as a very good practice of identifying what kind of rules we will use to get the individual derivatives , since there are all mixed up here compared to previous exercises.

Examples

$x=t^{2},\; y=2t$ $\frac{dy}{dx}=\frac{1}{t}$
Analyze : Parametric function. Parameter is

$x=\frac{1}{t},\; y=\left(t-2\right)^{5}$ $\frac{dy}{dx}=-5t^{2}\left(t-2\right)^{4}$
Analyze :
- $\begin{array}{ll} \cfrac{dx}{dt}=-\cfrac{1}{t^{2}}\qquad&\cfrac{dy}{dt}=5\left(t-2\right)^{4} \end{array}$
- $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$
- $={\color{Red}5\left(t-2\right)^{4}}$ $=5\left(t-2\right)^{4}{\color{Red}\times-t^{2}}$
- $=-5t^{2}\left(t-2\right)^{4}$

$x=\sin \theta,\; y=\cos \theta\;$ $\frac{dy}{dx}=-\frac{\sin \theta}{\cos \theta}$
Analyze : Parametric function. Parameter is $\theta\,$
- $\begin{array}{ll} \cfrac{dx}{d{\color{Red}\theta}}= \end{array}$
- $\begin{array}{ll} \cfrac{dx}{d\theta}=\cos \theta\qquad&\cfrac{dy}{d\theta}=-\sin \theta \end{array}$
- $\frac{dy}{dx}=\frac{dy}{d\theta}\times\frac{d\theta}{dx}$
- $=-\frac{\sin \theta}{\cos \theta}$

$x=\sin^{3}5t,\; y=\cos^{3}5t$ $\frac{dy}{dx}=-\frac{\cos 5 t}{\sin 5t}$
Analyze Both $x,y\,$ are something power n
- Rewrite it first if needed
- $x=\left(\sin 5t\right)^{3}\qquad\qquad\qquad\qquad y=\left(\cos 5t\right)^{3}$
- $\begin{array}{ll} \cfrac{dx}{dt}=3\left(\sin 5t\right)^{2}\cdot 5\cos5t\qquad&\cfrac{dy}{dt}=3\left(\cos 5t\right)^{2}\cdot -5\sin5t \end{array}$
- Simplify first
- $\begin{array}{ll} \cfrac{dx}{dt}=15\left(\sin 5t\right)^{2}\cos5t\qquad\;&\cfrac{dy}{dt}=-15\left(\cos 5t\right)^{2}\sin5t \end{array}$
- $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$
- $=\frac{-15\left(\cos 5t\right)^{2}\sin5t}{15\left(\sin 5t\right)^{2}\cos5t}$
- $=-\frac{\cos 5 t}{\sin 5t}$

$x=e^{3t}+2t,\; y=3-\ln t$ $\frac{dy}{dx}=-\frac{1}{t\left(3e^{3t}+2\right)}$
and are very different. Be sure to check the derivative is correct.
- $\begin{array}{ll} {\color{Red}\cfrac{dx}{dt}=3e^{3t}+2} \end{array}$
- $\begin{array}{ll} \cfrac{dx}{dt}=3e^{3t}+2\qquad&{\color{Red}\cfrac{dy}{dt}=-\cfrac{1}{t}} \end{array}$
- $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$
- $={\color{Red}-\cfrac{1}{t}}$ $=-\frac{1}{t{\color{Red}\left(3e^{3t}+2\right)}}$

$x=te^{t},\; y=\frac{e^{t}}{t}$ $\frac{dy}{dx}=\frac{t-1}{t^{2}\left(t+1\right)}$
Analyze : is
- $\begin{array}{ll} {\color{Red}\cfrac{dx}{dt}=te^{t}+e^{t}} \end{array}$
- $\begin{array}{ll} \cfrac{dx}{dt}=te^{t}+e^{t}\qquad&{\color{Red}\cfrac{dy}{dt}=\cfrac{te^{t}-e^{t}}{t^{2}}} \end{array}$
- Simplify first
- $\begin{array}{ll} \cfrac{dx}{dt}=e^{t}\left(t+1\right)\qquad&\cfrac{dy}{dt}=\cfrac{e^{t}\left(t-1\right)}{t^{2}} \end{array}$
- $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$
- ${\color{Red}=\frac{e^{t}\left(t-1\right)}{t^{2}}}$ $=\frac{e^{t}\left(t-1\right)}{t^{2}{\color{Red}e^{t}\left(t+1\right)}}$
- $\frac{dy}{dx}=\frac{t-1}{t^{2}\left(t+1\right)}$

Exercise 7

Find $\frac{dy}{dx}$ for the following

a) $x=4t^{2}-1,\;y=t^{3}-2t$ $\frac{dy}{dx}=\frac{3t^{2}-2}{8t}$
- $\begin{array}{ll} x=4t^{2}-1 & y=t^{3}-2t\\ \cfrac{dx}{dt}=8t & \cfrac{dy}{dt}=3t^{2}-2\\ \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=\frac{3t^{2}-2}{8t} \end{align}$

b) $x=3\left(t^{2}-3\right)^{6},\;y=1-t^{3}$ $\frac{dy}{dx}=-\frac{t}{12\left(t^{2}-3\right)^{5}}$
- $\begin{array}{ll} x=3\left(t^{2}-3\right)^{6} & y=1-t^{3}\\ \cfrac{dx}{dt}=18\left(t^{2}-3\right)^{5}\cdot 2t & \cfrac{dy}{dt}=-3t^{2}\\ \quad\;=36t\left(t^{2}-3\right)^{5} \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=\frac{-3t^{2}}{36t\left(t^{2}-3\right)^{5}}\\ &=-\frac{t}{12\left(t^{2}-3\right)^{5}} \end{align}$

c) $x=t-\frac{1}{t},\;y=t+\frac{1}{t}$ $\frac{dy}{dx}=\frac{t^{2}-1}{t^{2}+1}$
- $\begin{array}{ll} x=t-\cfrac{1}{t} & y=t+\cfrac{1}{t}\\ \cfrac{dx}{dt}=1+\cfrac{1}{t^{2}} & \cfrac{dy}{dt}=1-\cfrac{1}{t^{2}}\\ \quad\;=\cfrac{t^{2}+1}{t^{2}} & \quad\;=\cfrac{t^{2}-1}{t^{2}}\\ \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=\frac{t^{2}-1}{t^{2}}\times \frac{t^{2}}{t^{2}+1}\\ &=\frac{t^{2}-1}{t^{2}+1} \end{align}$

d) $x=\sqrt{t},\;y=\frac{3}{t^{3}-1}$ $\frac{dy}{dx}=-\frac{18t^{\frac{3}{2}}}{\left(t^{3}-1\right)^{2}}$
- $\begin{array}{ll} x=\sqrt{t} & y=\cfrac{3}{t^{3}-1}\\ \cfrac{dx}{dt}=\cfrac{1}{2\sqrt{t}} & \cfrac{dy}{dt}=-\cfrac{9t^{2}}{\left(t^{3}-1\right)^{2}}\\ \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=-\frac{9t^{2}}{\left(t^{3}-1\right)^{2}}\times2\sqrt{t}\\ &=-\frac{18t^{\frac{3}{2}}}{\left(t^{3}-1\right)^{2}}\\ \end{align}$

e) $x=3\cos t,\;y=3\sin t$ $\frac{dy}{dx}=-\frac{\cos t}{\sin t}$
- $\begin{array}{ll} x=3\cos t & y=3\sin t\\ \cfrac{dx}{dt}=-3\sin t & \cfrac{dy}{dt}=3\cos t\\ \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=\frac{3\cos t}{-3\sin t}\\ &=-\frac{\cos t}{\sin t} \end{align}$

f) $x=\sin^{4}2\theta ,\;y=\cos^{4}2\theta =$ $\frac{dy}{dx}=-\frac{\cos^{2}2\theta }{\sin^{2}2\theta }$
- $\begin{array}{ll} x=\sin^{4}2\theta & y=\cos^{4}2\theta \\ \cfrac{dx}{d\theta }=4\sin^{3}2\theta \cdot 2\cos 2\theta & \cfrac{dy}{d\theta }=4\cos^{3}2\theta \cdot -2\sin 2\theta \\ \quad\;=8\theta \sin^{3}2\theta \cos 2\theta & \quad\;=-8\theta \cos^{3}2\theta \sin 2\theta \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{d\theta}\times\frac{d\theta}{dx}\\ &=\frac{-8\theta\cos^{3}2\theta \sin 2\theta }{8\theta \sin^{3}2\theta \cos 2\theta }\\ &=-\frac{\cos^{2}2\theta }{\sin^{2}2\theta } \end{align}$

g) $x=\ln\left(t^{2}+1\right),\;y=\ln\left(e^{t}\sin 3t\right)$ $\frac{dy}{dx}=\frac{\left(t^{2}+1\right)\left(\sin 3t+3\cos 3t\right)}{2t\sin 3t}$
- $\begin{array}{ll} x=\ln\left(t^{2}+1\right) & y=\ln\left(e^{t}\sin 3t\right)\\ & \quad=\ln\left(e^{t}\right)+\ln\left(\sin 3t\right)\\ & \quad=t+\ln\left(\sin 3t\right)\\ \cfrac{dx}{dt}=\cfrac{2t}{t^{2}+1} & \cfrac{dy}{dt}=1+\cfrac{3\cos 3t}{\sin 3t}\\ & \quad\;=\cfrac{\sin 3t+3\cos 3t}{\sin 3t} \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=\frac{\sin 3t+3\cos 3t}{\sin 3t}\times\frac{t^{2}+1}{2t}\\ &=\frac{\left(t^{2}+1\right)\left(\sin 3t+3\cos 3t\right)}{2t\sin 3t} \end{align}$

h) $x=\frac{e^{t}-e^{-t}}{e^{2t}},\;y=\ln\sqrt{\frac{5}{t^{5}}}$ $\frac{dy}{dx}=-\frac{5}{2t\left(-e^{-t}+3e^{-3t}\right)}$
- $\begin{array}{ll} x=\cfrac{e^{t}-e^{-t}}{e^{2t}} & y=\ln\sqrt{\cfrac{5}{t^{5}}}\\ \quad=e^{-t}-e^{-3t} &\quad=\ln 5-\cfrac{5}{2}\ln t\\ \cfrac{dx}{dt}= -e^{-t}+3e^{-3t} & \cfrac{dy}{dt}=-\cfrac{5}{2t}\\ \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=-\frac{5}{2t\left(-e^{-t}+3e^{-3t}\right)}\\ \end{align}$

i) $x=t^{2}e^{2t},\;y=\frac{e^{2t}}{t^{2}}$ $\frac{dy}{dx}=\frac{\left(t-1\right)}{t^{4}\left(t+1\right)}$
- $\begin{array}{ll} x=t^{2}e^{2t} & y=\cfrac{e^{2t}}{t^{2}}\\ \cfrac{dx}{dt}=t^{2}\cdot 2e^{2t} +e^{2t} \cdot2t & \cfrac{dy}{dt}=\cfrac{t^{2}\cdot2e^{2t}-e^{2t}\cdot 2t}{t^{4}}\\ \cfrac{dx}{dt}=2te^{2t}\left(t+1\right) & \cfrac{dy}{dt}=\cfrac{2te^{2t}\left(t-1\right)}{t^{4}}\\ \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=\frac{2te^{2t}\left(t-1\right)}{t^{4}\cdot2te^{2t}\left(t+1\right)}\\ &=\frac{\left(t-1\right)}{t^{4}\left(t+1\right)} \end{align}$

j) $x=\sqrt{e^{t}}+\frac{1}{\sqrt{e^{t}}},\;y=\sqrt{e^{t}+4}$ $\frac{dy}{dx}=\frac{e^{\frac{3}{2}t}}{\sqrt{e^{t}+4}\left(e^{t}-1\right)}$
- $\begin{array}{ll} x=\sqrt{e^{t}}+\cfrac{1}{\sqrt{e^{t}}} & y=\sqrt{e^{t}+4}\\ \cfrac{dx}{dt}=\cfrac{1}{2}\sqrt{e^{t}}-\cfrac{1}{2\sqrt{e^{t}}} & \cfrac{dy}{dt}=\cfrac{e^{t}}{2\sqrt{e^{t}+4}}\\ \quad\;=\cfrac{e^{t}-1}{2\sqrt{e^{t}}} \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=\frac{e^{t}}{2\sqrt{e^{t}+4}}\times \cfrac{2\sqrt{e^{t}}}{e^{t}-1}\\ &=\frac{e^{\frac{3}{2}t}}{\sqrt{e^{t}+4}\left(e^{t}-1\right)} \end{align}$

k) $x=\cos\left(e^{4t}\right),\;y=\sin\left(e^{4t}\right)$ $\frac{dy}{dx}=-\frac{\cos\left(e^{4t}\right)}{\sin\left(e^{4t}\right)}$
- $\begin{array}{ll} x=\cos\left(e^{4t}\right) & y=\sin\left(e^{4t}\right)\\ \cfrac{dx}{dt}=-4e^{4t}\sin\left(e^{4t}\right) & \cfrac{dy}{dt}=4e^{4t}\cos\left(e^{4t}\right)\\ \end{array}$
- $\begin{align} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx}\\ &=\frac{4e^{4t}\cos\left(e^{4t}\right)}{-4e^{4t}\sin\left(e^{4t}\right)}\\ &=-\frac{\cos\left(e^{4t}\right)}{\sin\left(e^{4t}\right)}\\ \end{align}$

Implicit Functions

Up to now, we have dealt with functions which is given in the form of $y=f(x)\,$ , which are explicit functions.

Examples : $y=x^{2}, y=\ln \left( \sin 3x \right)$

Implicit functions are simply functions/relations NOT written explicitly in the form of $y=f(x)\,$ .

Examples : $y^{2}+x^{2}=x, \frac{1}{y}+\frac{1}{x}=y, \sin y =x^{2}$

Note :

It is normally (but not always) difficult/impossible to rearrange the given implicit functions/relations to the explicit form (so don't waste your time)
There are however certain times we would instead change an explicit function to the implicit function to make the differentiation easier.
The derivative of the implicit function will be in terms of both $x\,$ and $y\,$ . This is perfectly acceptable if the question was given in the implicit form, unless they ask us to change it back to in terms of $x\,$ only, or the question was actually in explicit form but we used implicit.

Derivative of f(y)

If have $f\left({\color{Red}x}\right)$ , we can find $\frac{d\left[f\left({\color{Red}x}\right)\right]}{d{\color{Red}x}}$ directly.
If we have $f\left({\color{Blue}y}\right)$ however, we can only find

$\frac{d\left[f\left({\color{Blue}y}\right)\right]}{d{\color{Red}x}}$ $=\frac{d\left[f\left({\color{Blue}y}\right)\right]}{?}\times\frac{?}{d{\color{Red}x}}$

$=\frac{d\left[f\left({\color{Blue}y}\right)\right]}{{\color{Blue}dy}}\times\frac{{\color{Blue}dy}}{d{\color{Red}x}}$

In other words

Multiply $\frac{dy}{dx}$

Once again

$\frac{d\left[f\left(y\right)\right]}{dx}$ $=\frac{d\left[f\left(y\right)\right]}{dy}\times\frac{dy}{dx}$

Examples

$=2y\frac{dy}{dx}$
- Or more quickly
  - Differentiate $f\left(y\right)$ ${\color{Red}2y}$
  - Multiply $\frac{dy}{dx}$ $2y{\color{Red}\frac{dy}{dx}}$

$\frac{d}{dx}\left(y^{3}\right)$ $=3y^{2}{\color{Red}\frac{dy}{dx}}$
$\frac{d}{dx}\left(3y\right)$ $=3{\color{Red}\frac{dy}{dx}}$ (Don't forget the $\frac{dy}{dx}$ )
$\frac{d}{dx}\left(\frac{1}{y}\right)$ $=-\frac{1}{y^2}\frac{dy}{dx}$
$\frac{d}{dx}\left(\sqrt{y}\right)$ $=\frac{1}{2\sqrt{y}}\frac{dy}{dx}$
$\frac{d}{dx}\left(e^{y}\right)$ $=e^{y}\frac{dy}{dx}$
$\frac{d}{dx}\left(\ln y\right)$ $=\frac{1}{y}\frac{dy}{dx}$
$\frac{d}{dx}\left(\sin y\right)$ $=\left(\cos y\right)\frac{dy}{dx}$

Products/Quotients

As before, basically, just remember that products/quotients takes precedence (as in we apply the products/quotients formulas first before anything else). In fact, there is nothing new to learn here, just a practice of sticking to the formulas even when things get very complicated (and as always, learning how to check to make sure we didn't miss out anything)

Let's try

- Analyze :
  - Copy first thing $={\color{Red}x}$
  - Differentiate second thing $=x{\color{Red}\frac{dy}{dx}}$
  - Plus copy second thing $=x\frac{dy}{dx}{\color{Red}+y}$

$=\frac{x\cfrac{dy}{dx}-y}{x^2}$
- - $=\frac{\qquad\qquad}{}$
  - $=\frac{{\color{Red}x}\qquad\qquad}{}$
  - $=\frac{x{\color{Red}\cfrac{dy}{dx}}\qquad\qquad}{}$
  - $=\frac{x\cfrac{dy}{dx}{\color{Red}-}}{}$
  - $=\frac{x\cfrac{dy}{dx}-{\color{Red}y}}{}$
  - $=\frac{x\cfrac{dy}{dx}-y}{{\color{Red}x^2}}$

$\frac{d}{dx}\left(yx^2\right)$ $=y\left(2x\right)+x^{2}\frac{dy}{dx}$

Examples

Differentiate the following w.r.t. $x\,$ . Hence. find $\frac{dy}{dx}$

$x^{2}+y^{2}=x\,$ $\frac{dy}{dx}=\frac{1-2x}{2y}$
Analyze :

$xy^{2}=\sin x\,$ $\frac{dy}{dx}=\frac{\cos x-y^{2}}{2xy}$
Analyze : Implicit function
- $\mbox{Diff. w.r.t.}\;x :$
  ${\color{Red}x}$
  $x{\color{Red}\cdot2y}$
  $x\cdot2y{\color{Red}\frac{dy}{dx}}$
  $x\cdot2y\frac{dy}{dx}{\color{Red}+y^{2}}$
  $x\cdot2y\frac{dy}{dx}+y^{2}{\color{Red}=\cos x}$
- Rearrange
  $2xy{\color{Blue}\frac{dy}{dx}}=\cos x-y^{2}$
  ${\color{Blue}\frac{dy}{dx}}=\frac{\cos x-y^{2}}{2xy}$

$\ln y+\sin y=\sin 3x\,$ $\frac{dy}{dx}=\frac{3y\cos3x}{1+y\cos y}$
Analyze : Implicit function

$\ln\left(x+y\right)=4x$ $\frac{dy}{dx}=4\left(x+y\right)-1$
Analyze :
- $\mbox{Diff. w.r.t.}\;x :$
  $\frac{\quad}{{\color{Red}x+y}}$
  $\frac{{\color{Red}1}\quad}{x+y}$
  $\frac{1{\color{Red}+\frac{dy}{dx}}}{x+y}$
  $\frac{1+\frac{dy}{dx}}{x+y}{\color{Red}=4}$
- $1+{\color{Blue}\frac{dy}{dx}}=4\left(x+y\right)$
- ${\color{Blue}\frac{dy}{dx}}=4\left(x+y\right)+1$

Exercise 8

Differentiate the following with respect to $x\,$ . Hence, find $\frac{dy}{dx}$ .

a) $\frac{dy}{dx}=\frac{x}{y}$
- $\begin{align} &x^{2}-y^{2}=5\\ &\mbox{Diff. w.r.t.}\;x :\\ &2x-2y\frac{dy}{dx}=0\\ &2y\frac{dy}{dx}=2x\\ &\frac{dy}{dx}=\frac{x}{y}\\ \end{align}$

b) $\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$
- $\begin{align} &\sqrt{x}+\sqrt{y}=2\\ &\mbox{Diff. w.r.t.}\;x :\\ &\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0\\ &\frac{1}{2\sqrt{y}}\frac{dy}{dx}=-\frac{1}{2\sqrt{x}}\\ &\frac{dy}{dx}=-\sqrt{\frac{y}{x}}\\ \end{align}$

c) $\frac{dy}{dx}=-\frac{\left(y+\cos x\right)}{x}$
- $\begin{align} &xy+\sin x=0\\ &\mbox{Diff. w.r.t.}\;x :\\ &x\frac{dy}{dx}+y+\cos x=0\\ &x\frac{dy}{dx}=-\left(y+\cos x\right)\\ &\frac{dy}{dx}=-\frac{\left(y+\cos x\right)}{x}\\ \end{align}$

d) $\frac{dy}{dx}=\frac{e^{x}-2xy^{3}}{3x^{2}y^{2}}$
- $\begin{align} &x^{2}y^{3}=e^{x}\\ &\mbox{Diff. w.r.t.}\;x :\\ &x^{2}\cdot3y^{2}\frac{dy}{dx}+y^{3}\left(2x\right)=e^{x}\\ &3x^{2}y^{2}\frac{dy}{dx}=e^{x}-2xy^{3}\\ &\frac{dy}{dx}=\frac{e^{x}-2xy^{3}}{3x^{2}y^{2}}\\ \end{align}$

e) $\frac{dy}{dx}=\frac{y^{2}}{x^{2}}$
- $\begin{align} &\frac{1}{x}-\frac{1}{y}=10\\ &\mbox{Diff. w.r.t.}\;x :\\ &-\frac{1}{x^{2}}+\frac{1}{y^{2}}\frac{dy}{dx}=0\\ &\frac{1}{y^{2}}\frac{dy}{dx}=\frac{1}{x^{2}}\\ &\frac{dy}{dx}=\frac{y^{2}}{x^{2}}\\ \end{align}$

f) $\frac{dy}{dx}=-\frac{\cos x}{\cos y}$
- $\begin{align} &\sin y+\sin x=1\\ &\mbox{Diff. w.r.t.}\;x :\\ &\left(\cos y\right)\frac{dy}{dx}+\cos x=0\\ &\left(\cos y\right)\frac{dy}{dx}=-\cos x\\ &\frac{dy}{dx}=-\frac{\cos x}{\cos y}\\ \end{align}$

g) $\frac{dy}{dx}=4\left(y+4\right)$
- $\begin{align} &\ln \left(y+4\right)=4x\\ &\mbox{Diff. w.r.t.}\;x :\\ &\frac{1}{y+4}\frac{dy}{dx}=4\\ &\frac{dy}{dx}=4\left(y+4\right)\\ \end{align}$

h) $\frac{dy}{dx}=\frac{2\cos 2x}{e^{x+y}}-1$
- $\begin{align} &e^{x+y}=\sin 2x\\ &\mbox{Diff. w.r.t.}\;x :\\ &\left(1+\frac{dy}{dx}\right)e^{x+y}=2\cos 2x\\ &1+\frac{dy}{dx}=\frac{2\cos 2x}{e^{x+y}}\\ &\frac{dy}{dx}=\frac{2\cos 2x}{e^{x+y}}-1\\ \end{align}$

i) $\frac{dy}{dx}=\frac{1-y\cos\left(xy\right)}{x\cos\left(xy\right)}$
- $\begin{align} &\sin\left(xy\right)=x\\ &\mbox{Diff. w.r.t.}\;x :\\ &\left(x\frac{dy}{dx}+y\right)\cos\left(xy\right)=1\\ &\left(x\frac{dy}{dx}+y\right)=\frac{1}{\cos\left(xy\right)}\\ &x\frac{dy}{dx}=\frac{1}{\cos\left(xy\right)}-y\\ &\frac{dy}{dx}=\frac{1-y\cos\left(xy\right)}{x\cos\left(xy\right)}\\ \end{align}$

j) $\frac{dy}{dx}=\frac{xe^{x^{2}}}{\sin y \cos y}$
- $\begin{align} &\sin^{2}y=e^{x^{2}}\\ &\mbox{Diff. w.r.t.}\;x :\\ &2\sin y \cos y\frac{dy}{dx}=2xe^{x^{2}}\\ &\frac{dy}{dx}=\frac{xe^{x^{2}}}{\sin y \cos y} \end{align}$

Derivatives using logarithm

Prove of $\frac{d}{dx}{a}^{x} ={a}^{x}\ln a \,$

Let $y={a}^{x}\,$
$\ln y=\ln{a}^{x}\,$
$\ln y=x\ln{a}\,$
$\frac{1}{y}\frac{dy}{dx}=\ln a$
$\frac{dy}{dx}=y\ln a$
$\frac{dy}{dx}={a}^{x}\ln a$

Note: Alternatively, we can prove this formula using the identity $e^{\ln x}=x\,$ and $\frac{d}{dx}\left(e^{x}\right)=e^{x}$

Notice that the method works because logarihms allows us to bring down the index to make the differentiating much easier. Additionally, we can also use it to change times/divide to plus/minus.

Thus, we can use this method to as alternative method to solve questions with difficult index, such as $\frac{3x-2}{\sqrt[3]{x+1}}$ (actually, the differentiation itself isn't that difficult even if we use the normal method, but the simplifying after that can be tricky)

Note:

Remember to ln BOTH sides
Remember the $\frac{dy}{dx}$
Remember to change the final answer back to $x\,$ only

Example

Using logarithm, or otherwise, differentiate $y=\frac{3x-2}{\sqrt[3]{x+1}}$ w.r.t. $x\,$

BOTH side $\ln y=\ln \left[\frac{3x-2}{\sqrt[3]{x+1}}\right]$
- Differentiate
  ${\color{Red}\frac{1}{y}}$
  $\frac{1}{y}{\color{Red}\frac{dy}{dx}}$
  $\frac{1}{y}\frac{dy}{dx}={\color{Red}\frac{}{3x-2}}$
  $\frac{1}{y}\frac{dy}{dx}=\frac{{\color{Red}3}}{3x-2}$
  $\frac{1}{y}\frac{dy}{dx}=\frac{3}{3x-2}{{\color{Red}-\frac{1}{3}}}$
  $\frac{1}{y}\frac{dy}{dx}=\frac{3}{3x-2}-\frac{1}{3{\color{Red}\left(x+1\right)}}$
- $\frac{1}{y}\frac{dy}{dx}=\frac{9\left(x+1\right)-\left(3x-2\right)}{3\left(3x-2\right)\left(x+1\right)}$
- $\frac{1}{y}\frac{dy}{dx}=\frac{6x+11}{3\left(3x-2\right)\left(x+1\right)}$
- $\frac{dy}{dx}=\left(\frac{3x-2}{\sqrt[3]{x+1}}\right)\frac{6x+11}{3\left(3x-2\right)\left(x+1\right)}$
- Simplify (be careful)
- $\frac{dy}{dx}=\frac{6x+11}{3\left(x+1\right)^{\frac{4}{3}}}$

Alternatively,

$\begin{align} y&=\frac{3x-2}{\sqrt[3]{x+1}}\\ \frac{dy}{dx}&=\frac{\left(x+1\right)^{\frac{1}{3}}\cdot3-\left(3x-2\right)\cdot\frac{1}{3}\left(x+1\right)^{-\frac{2}{3}}}{\left(x+1\right)^{\frac{2}{3}}}\\ &=\frac{\left(x+1\right)^{-\frac{2}{3}}\left[9\left(x+1\right)-\left(3x-2\right)\right]}{3\left(x+1\right)^{\frac{2}{3}}}\\ &=\frac{6x+11}{3\left(x+1\right)^{\frac{4}{3}}}\\ \end{align}$

Exercise 9

Using logarithm, or otherwise, find $\frac{dy}{dx}$ for the following.

$\frac{dy}{dx}=\frac{\left(30x-17\right)}{4\left(3x-1\right)^{\frac{3}{4}}}$
- $\begin{align} &y=\sqrt[4]{3x-1}\left(2x-3\right)\\ &\ln y= \ln \left[ \sqrt[4]{3x-1}\left(2x-3\right)\right]\\ &\ln y= \frac{1}{4}\ln\left(3x-1\right)+\ln \left(2x-3\right)\\ &\frac{1}{y}\frac{dy}{dx}= \frac{3}{4\left(3x-1\right)}+\frac{2}{\left(2x-3\right)}\\ &\frac{1}{y}\frac{dy}{dx}= \frac{3\left(2x-3\right)+8\left(3x-1\right)}{4\left(3x-1\right)\left(2x-3\right)}\\ &\frac{1}{y}\frac{dy}{dx}= \frac{30x-17}{4\left(3x-1\right)\left(2x-3\right)}\\ &\frac{dy}{dx}= y\left[\frac{30x-17}{4\left(3x-1\right)\left(2x-3\right)}\right]\\ &\frac{dy}{dx}= \sqrt[4]{3x-1}\left(2x-3\right)\left[\frac{30x-17}{4\left(3x-1\right)\left(2x-3\right)}\right]\\ &\frac{dy}{dx}= \frac{30x-17}{4\left(3x-1\right)^{\frac{3}{4}}}\\ \end{align}$
- $\begin{align} y&=\sqrt[4]{3x-1}\left(2x-3\right)\\ \frac{dy}{dx}&=\left(3x-1\right)^{\frac{1}{4}}\cdot2+\left(2x-3\right)\cdot\frac{1}{4}\left(3x-1\right)^{-\frac{3}{4}}\left(3\right)\\ &=\frac{1}{4}\left(3x-1\right)^{-\frac{3}{4}}\left[8\left(3x-1\right)+3\left(2x-3\right) \right]\\ &=\frac{30x-17}{4\left(3x-1\right)^{\frac{3}{4}}}\\ \end{align}$

$\frac{dy}{dx}=\frac{\left(11x+13\right)\left(x-1\right)^{3}}{3\left(x+1\right)^{\frac{4}{3}}}$
- $\begin{align} &y=\frac{\left(x-1\right)^{4}}{\sqrt[3]{x+1}}\\ &\ln y= \ln \left[\frac{\left(x-1\right)^{4}}{\sqrt[3]{x+1}}\right]\\ &\ln y= 4\ln\left(x-1\right)-\frac{1}{3}\ln \left(x+1\right)\\ &\frac{1}{y}\frac{dy}{dx}= \frac{4}{\left(x-1\right)}-\frac{1}{3\left(x+1\right)}\\ &\frac{1}{y}\frac{dy}{dx}= \frac{12\left(x+1\right)-\left(x-1\right)}{3\left(x-1\right)\left(x+1\right)}\\ &\frac{1}{y}\frac{dy}{dx}= \frac{11x+13}{3\left(x-1\right)\left(x+1\right)}\\ &\frac{dy}{dx}= y\left[\frac{11x+13}{3\left(x-1\right)\left(x+1\right)}\right]\\ &\frac{dy}{dx}= \frac{\left(x-1\right)^{4}}{\sqrt[3]{x+1}}\left[\frac{11x+13}{3\left(x-1\right)\left(x+1\right)}\right]\\ &\frac{dy}{dx}= \frac{\left(11x+13\right)\left(x-1\right)^{3}}{3\left(x+1\right)^{\frac{4}{3}}}\\ \end{align}$
- $\begin{align} y&=\frac{\left(x-1\right)^{4}}{\sqrt[3]{x+1}}\\ \frac{dy}{dx}&=\frac{\left(x+1\right)^{\frac{1}{3}}\cdot4\left(x-1\right)^{3}-\left(x-1\right)^{4}\cdot\frac{1}{3}\left(x+1\right)^{-\frac{2}{3}}}{\left(x+1\right)^{\frac{2}{3}}}\\ \frac{dy}{dx}&=\frac{\left(x+1\right)^{-\frac{2}{3}}\left(x-1\right)^{3}\left[12\left(x+1\right)-\left(x-1\right)\right]}{3\left(x+1\right)^{\frac{2}{3}}}\\ \frac{dy}{dx}&= \frac{\left(11x+13\right)\left(x-1\right)^{3}}{3\left(x+1\right)^{\frac{4}{3}}}\\ \end{align}$

Higher Derivatives

If we differentiate $\frac{dy}{dx}$ w.r.t. $x\,$ , what do we get?

$\frac{d}{dx}\left(\frac{dy}{dx}\right)$ $=\frac{d^{2}y}{dx^{2}}$
Note : $\frac{d^{2}y}{dx^{2}} \neq \left(\frac{dy}{dx}\right)^{2}$

If we keep on differentiation, the notation will be as follows

$y \rightarrow$ $\frac{dy}{dx} \rightarrow$ $\frac{d^{2}y}{dx^{2}}\rightarrow$ $\frac{d^{3}y}{dx^{3}}\rightarrow \ldots$
$f\left(x\right) \rightarrow$

Example

$y=x^{5}\,$
$\frac{dy}{dx}$ $=5x^{4}\,$
$\frac{d^{2}y}{dx^{2}}$ $=20x^{3}\,$
$\frac{d^{3}y}{dx^{3}}$ $=60x^{2}\,$

Proving Differential Equations

These are a special type of question whereby we are given a function, and instead of finding the actual formula for its derivative(s), we are asked to prove that the function satisfies a given differential equation.

Example : If $y=e^{-x}\left(ax+b\right)$ , where $a, b\,$ are constants, prove that $\frac{d^{2}y}{dx^2}+2\frac{dy}{dx}+y=0$

Notice that :
- We are NOT required to actually find the formula of the derivatives (in terms of $x\,$ anyway)
- The final equation doesn't have most of the terms in the original function
- One might say, why not just differentiate it twice, and plug it into the L.H.S and prove it to be zero? Trust me when I say it's not going to work, not just because it might not get accepted (I am not sure), but because it will be so complicated that you will most likely make many careless mistake and waste a lot of time.

General Guidelines :

Each question, is of course, different. But always keep the following two tips in mind when attempting this kind of question

If needed, rearrange the equation BEFORE differentiating.
- This doesn't only apply to the first step, but also before you differentiate it a second time. Since we are not interested in keeping the derivative in terms of $x\,$ only, we have more freedom here compared to usual. We can bring over terms by multiplying, square both sides, etc.
Always simplify if possible BEFORE differentiating it a second time.
- "Simplify" here carry a different meaning than usual. As most of the time the final equation contains a lot of $y\,$ 's ( $y, \frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}$ ), it will mean we try to "simplify" it so that it contains those terms, rather than complicated terms in $x\,$

Examples

If $y=e^{-x}\left(ax+b\right)$ , where $a, b\,$ are constants, prove that $\frac{d^{2}y}{dx^2}+2\frac{dy}{dx}+y=0$
The first thing we do is look at the required equation
But for now we can't see what else we can do but differentiate it, so let's do it the first time
- $y=e^{-x}\left(ax+b\right)$
- $\frac{dy}{dx}=e^{-x}a-\left(ax+b\right)e^{-x}$

So what should we do then? Let's look at what we have now
- $\begin{align} y&=e^{-x}\left(ax+b\right)\\ \frac{dy}{dx}&=e^{-x}a-\left(ax+b\right)e^{-x} \end{align}$ See something?
- $\begin{align} y&={\color{Red}e^{-x}\left(ax+b\right)}\\ \frac{dy}{dx}&=e^{-x}a-{\color{Red}\left(ax+b\right)e^{-x}} \end{align}$
- $\begin{align} y&=e^{-x}\left(ax+b\right)\\ \frac{dy}{dx}&=e^{-x}a-\left(ax+b\right)e^{-x}\\ \frac{dy}{dx}&=ae^{-x}-y\\ \end{align}$

Seems like nothing else we can do so let's differentiate it a second time
- $\frac{d^{2}y}{dx^{2}}=-ae^{-x}-\frac{dy}{dx}$

So lets look at what we have and we we need
- $\frac{d^{2}y}{dx^{2}}=-ae^{-x}-\frac{dy}{dx}$
- $\frac{d^{2}y}{dx^2}+2\frac{dy}{dx}+y=0$
- So lets see what we have extra
  $\frac{d^{2}y}{dx^{2}}={\color{Red}-ae^{-x}}-\frac{dy}{dx}$
  $\frac{d^{2}y}{dx^2}+2\frac{dy}{dx}+y=0$
But substitute with what? Let's look at the working thus far again
- $\begin{align} y&=e^{-x}\left(ax+b\right)\\ \frac{dy}{dx}&=e^{-x}a-\left(ax+b\right)e^{-x}\\ \frac{dy}{dx}&=ae^{-x}-y\\ \frac{d^{2}y}{dx^{2}}&=-ae^{-x}-\frac{dy}{dx} \end{align}$

Completing our working
- $\begin{align} y&=e^{-x}\left(ax+b\right)\\ \frac{dy}{dx}&=e^{-x}a-\left(ax+b\right)e^{-x}\\ \frac{dy}{dx}&=ae^{-x}-y \qquad\frac{\qquad\qquad}{}\left(1\right)\\ \frac{d^{2}y}{dx^{2}}&=-ae^{-x}-\frac{dy}{dx}\qquad\frac{\qquad\qquad}{}\left(2\right) \end{align}$
- $\mbox{From}\;\left(1\right) : ae^{-x}=\frac{dy}{dx}+y\qquad\frac{\qquad\qquad}{}\left(3\right)$
- $\left(3\right)\rightarrow \left(2\right): \frac{d^{2}y}{dx^{2}}=-\left(\frac{dy}{dx}+y\right)-\frac{dy}{dx}$
- $\therefore \frac{d^{2}y}{dx^2}+2\frac{dy}{dx}+y=0\qquad\frac{\qquad\qquad}{}\mbox{Proved}$

That was quite difficult. Is there a better way? Especially since we see the final equation seems so simple.

Differentiate
- ${\color{Red}y}$
- $y{\color{Red}e^{x}}$
- $ye^{x}{\color{Red}+\frac{dy}{dx}}$
- $ye^{x}+\frac{dy}{dx}{\color{Red}e^{x}}$
- $ye^{x}+\frac{dy}{dx}e^{x}={\color{Red}a}$

Before differentiating the second time, we simplify
- $e^{x}\left(y+\frac{dy}{dx}\right)=a$

Differentiate carefully
- ${\color{Red}e^{x}}$
- $e^{x}{\color{Red}\left(\frac{dy}{dx}\right.}$
- $e^{x}\left(\frac{dy}{dx}+{\color{Red}\frac{d^{2}y}{dx^{2}}}\right)$
- $e^{x}\left(\frac{dy}{dx}+\frac{d^{2}y}{dx^{2}}\right)+{\color{Red}\left(y+\frac{dy}{dx}\right)}$
- $e^{x}\left(\frac{dy}{dx}+\frac{d^{2}y}{dx^{2}}\right)+\left(y+\frac{dy}{dx}\right){\color{Red}e^{x}}$
- $e^{x}\left(\frac{dy}{dx}+\frac{d^{2}y}{dx^{2}}\right)+\left(y+\frac{dy}{dx}\right)e^{x}{\color{Red}=0}$

Completing the working
- $e^{x}\left(\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+y\right)=0$
- $\mbox{Since}\; e^{x}>0, \therefore\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+y=0\qquad\frac{\qquad\qquad}{}\mbox{Proved}$

If $y=e^{2x}\sin 3x\,$ , prove that $\frac{d^{2}y}{dx^2}-4\frac{dy}{dx}+13y=0$
The first step would be to rearrange the equation
- $y=e^{2x}\sin 3x\,$
- $\therefore ye^{-2x}=\sin 3x$
- $\mbox{Diff. w.r.t.}\;x :$
  ${\color{Red}y}$
  $y{\color{Red}\left(-2e^{-2x}\right)}$
  $y\left(-2e^{-2x}\right){\color{Red}+e^{-2x}}$
  $y\left(-2e^{-2x}\right)+e^{-2x}{\color{Red}\frac{dy}{dx}}$
  $y\left(-2e^{-2x}\right)+e^{-2x}\frac{dy}{dx}={\color{Red}\quad\cos 3x}$
  $y\left(-2e^{-2x}\right)+e^{-2x}\frac{dy}{dx}={\color{Red}3}\cos 3x$
- $e^{-2x}\left(\frac{dy}{dx}-2y\right)=3\cos 3x$
- $\mbox{Diff. w.r.t.}\;x :$
  ${\color{Red}e^{-2x}}$
  $e^{-2x}{\color{Red}\left(\frac{d^{2}y}{dx^{2}}\right.}$
  $e^{-2x}\left(\frac{d^{2}y}{dx^{2}}{\color{Red}-2\frac{dy}{dx}}\right)$
  $e^{-2x}\left(\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}\right)+{\color{Red}\left(\frac{dy}{dx}-2y\right)}$
  $e^{-2x}\left(\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}\right)+\left(\frac{dy}{dx}-2y\right){\color{Red}\left(-2e^{-2x}\right)}$
  $e^{-2x}\left(\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}\right)+\left(\frac{dy}{dx}-2y\right)\left(-2e^{-2x}\right)={\color{Red}-9\sin 3x}$
- $e^{-2x}\left[\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}-2\left(\frac{dy}{dx}-2y\right)\right]=-9\sin 3x$
- $\frac{d^{2}y}{dx^{2}}-4\frac{dy}{dx}+4y=-9e^{2x}\sin 3x$
- $\frac{d^{2}y}{dx^{2}}-4\frac{dy}{dx}+4y=-9y$
- $\frac{d^{2}y}{dx^{2}}-4\frac{dy}{dx}+13y=0\qquad\frac{\qquad\qquad}{}\mbox{Proved}$

If $y=\sqrt{\sin x}$ , prove that $2y\left(\frac{d^{2}y}{dx^2}\right)+2\left(\frac{dy}{dx}\right)^{2}+y^{2}=0$
The first step would be to
- $y=\sqrt{\sin x}$
- $\therefore y^{2}=\sin x$
- $\mbox{Diff. w.r.t.}\;x :$
  ${\color{Red}2y}$
  $2y{\color{Red}\frac{dy}{dx}}$
  $2y\frac{dy}{dx}={\color{Red}\cos x}$
- $\mbox{Diff. w.r.t.}\;x :$
  ${\color{Red}2y}$
  $2y{\color{Red}\frac{d^{2}y}{dx^{2}}}$
  $2y\frac{d^{2}y}{dx^{2}}+{\color{Red}\frac{dy}{dx}}$
  $2y\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}{\color{Red}\cdot 2\frac{dy}{dx}}$
  $2y\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\cdot 2\frac{dy}{dx}={\color{Red}-\sin x}$
- $2y\left(\frac{d^{2}y}{dx^2}\right)+2\left(\frac{dy}{dx}\right)^{2}=-y^{2}$
- $2y\left(\frac{d^{2}y}{dx^2}\right)+2\left(\frac{dy}{dx}\right)^{2}+y^{2}=0\qquad\frac{\qquad\qquad}{}\mbox{Proved}$

If $y=\ln{\left(\sin x\right)}$ , prove that $\frac{d^{2}y}{dx^2}+\left(\frac{dy}{dx}\right)^{2}+1=0$
Analyze :
- $\mbox{Diff. w.r.t.}\;x :$
  ${\color{Red}\frac{dy}{dx}}$
  $\frac{dy}{dx}={\color{Red}\frac{}{\sin x}}$
  $\frac{dy}{dx}=\frac{{\color{Red}\cos x}}{\sin x}$
- $\sin x\frac{dy}{dx}=\cos x$
- $\mbox{Diff. w.r.t.}\;x :$
  ${\color{Red}\sin x}$
  $\left(\sin x \right){\color{Red}\frac{d^{2}y}{dx^{2}}}$
  $\left(\sin x \right)\frac{d^{2}y}{dx^{2}}{\color{Red}+\frac{dy}{dx}}$
  $\left(\sin x \right)\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}{\color{Red}\cos x}$
  $\left(\sin x \right)\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\cos x={\color{Red}-\sin x}$
- $\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\left(\frac{\cos x}{\sin x}\right)=-1$
- Completing the working
  $\begin{align} &y=\ln{\left(\sin x\right)}\\ &\frac{dy}{dx}=\frac{\cos x}{\sin x}\\ &\sin x\frac{dy}{dx}=\cos x\qquad\frac{\qquad\qquad}{}\left(1\right)\\ &\left(\sin x \right)\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\cos x=-\sin x\\ &\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\left(\frac{\cos x}{\sin x}\right)=-1\qquad\frac{\qquad\qquad}{}\left(2\right)\\ \end{align}$
  $\left(1\right)\rightarrow \left(2\right): \frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\left(\frac{dy}{dx}\right)=-1$
  $\therefore \frac{d^{2}y}{dx^2}+\left(\frac{dy}{dx}\right)^{2}+1=0\qquad\frac{\qquad\qquad}{}\mbox{Proved}$

Exercise 10

1)If , prove that
- $\begin{align} &y=e^{3x}\left(9x+1\right)\\ &ye^{-3x}=\left(9x+1\right)\\ &\mbox{Diff. w.r.t.}\;x :\\ &y\left(-3e^{-3x}\right)+e^{-3x}\frac{dy}{dx}=9\\ &e^{-3x}\left(\frac{dy}{dx}-3y\right)=9\\ &\mbox{Diff. w.r.t.}\;x :\\ &e^{-3x}\left(\frac{d^{2}y}{dx^{2}}-3\frac{dy}{dx}\right)+\left(\frac{dy}{dx}-3y\right)\left(-3e^{-3x}\right)=0\\ &e^{-3x}\left(\frac{d^{2}y}{dx^{2}}-6\frac{dy}{dx}+9y\right)=0\\ &\mbox{Since}\;e^{-3x}>0, \therefore \frac{d^{2}y}{dx^{2}}-6\frac{dy}{dx}+9y\qquad\frac{\qquad\qquad}{}\mbox{Proved} \end{align}$

2)If , prove that
- $\begin{align} &y=\frac{\sin x}{x}\\ &xy=\sin x\\ &\mbox{Diff. w.r.t.}\;x :\\ &x\frac{dy}{dx}+y=\cos x\\ &\mbox{Diff. w.r.t.}\;x :\\ &x\left(\frac{d^{2}y}{dx^{2}}\right)+\frac{dy}{dx}+\frac{dy}{dx}=-\sin x\\ &x\left(\frac{d^{2}y}{dx^{2}}\right)+2\frac{dy}{dx}=-xy\\ &x\left(\frac{d^{2}y}{dx^{2}}\right)+2\frac{dx}{dy}+xy=0\qquad\frac{\qquad\qquad}{}\mbox{Proved}\\ \end{align}$
3)If , prove that
- $\begin{align} &y=\frac{1}{1+x^{2}}\\ &\left(1+x^{2}\right)y=1\\ &\mbox{Diff. w.r.t.}\;x :\\ &\left(1+x^{2}\right)\frac{dy}{dx}+y\left(2x\right)=0\\ &\mbox{Diff. w.r.t.}\;x :\\ &\left(1+x^{2}\right)\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\left(2x\right)+y\left(2\right)+\left(2x\right)\frac{dy}{dx}=0\\ &\left(1+x^{2}\right)\frac{d^{2}y}{dx^{2}}+4x\frac{dy}{dx}+2y=0\qquad\frac{\qquad\qquad}{}\mbox{Proved}\\\\ \end{align}$
4)If , prove that
- $\begin{align} &y=\ln\left(\sin x+\cos x\right)\\ &\mbox{Diff. w.r.t.}\;x :\\ &\frac{dy}{dx}=\frac{\cos x-\sin x}{\sin x+\cos x}\qquad\frac{\qquad\qquad}{}\left(1\right)\\ &\left(\sin x+\cos x\right)\frac{dy}{dx}=\left(\cos x-\sin x\right)\\ &\mbox{Diff. w.r.t.}\;x :\\ &\left(\sin x+\cos x\right)\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\left(\cos x-\sin x\right)=\left(-\sin x-\cos x\right)\\ &\left(\sin x+\cos x\right)\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\left(\cos x-\sin x\right)=-\left(\sin x+\cos x\right)\\ &\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\left(\frac{\cos x-\sin x}{\sin x+\cos x}\right)=-1\qquad\frac{\qquad\qquad}{}\left(2\right)\\ &\left(1\right)\rightarrow \left(2\right): \frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\left(\frac{dy}{dx}\right)=-1\\ &\frac{d^{2}y}{dx^{2}}+\left(\frac{dx}{dy}\right)^{2}+1=0\\ \end{align}$

5)If , where are constants, prove that
- $\begin{align} &ye^{mx}=a\cos px\qquad\frac{\qquad\qquad}{}\left(1\right)\\ &\mbox{Diff. w.r.t.}\;x :\\ &yme^{mx}+e^{mx}\frac{dy}{dx}=-ap\sin px\\ &e^{mx}\left(my+\frac{dy}{dx}\right)=-ap\sin px\\ &\mbox{Diff. w.r.t.}\;x :\\ &e^{mx}\left(m\frac{dy}{dx}+\frac{d^{2}y}{dx^{2}}\right)+\left(my+\frac{dy}{dx}\right)me^{mx}=-ap^{2}\cos px\\ &e^{mx}\left(m\frac{dy}{dx}+\frac{d^{2}y}{dx^{2}}+m^{2}y+m\frac{dy}{dx}\right)=-ap^{2}\cos px\\ &e^{mx}\left(\frac{d^{2}y}{dx^{2}}+2m\frac{dy}{dx}+m^{2}y\right)=-p^{2}\left(a\cos px\right)\qquad\frac{\qquad\qquad}{}\left(2\right)\\ &\left(1\right)\rightarrow \left(2\right):e^{mx}\left(\frac{d^{2}y}{dx^{2}}+2m\frac{dy}{dx}+m^{2}y\right)=-p^{2}\left(ye^{mx}\right)\\ &e^{mx}\left[\frac{d^{2}y}{dx^{2}}+2m\frac{dy}{dx}+\left( m^{2}+p^{2}\right)y\right]=0\\ &\mbox{Since}\;e^{mx}>0, \therefore \frac{d^{2}y}{dx^{2}}+2m\frac{dx}{dy}+\left(m^{2}+p^{2}\right)y=0\qquad\frac{\qquad\qquad}{}\mbox{Proved} \end{align}$