Differentiation Part2

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Chain Rule

$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

Notice that

Composite Functions

Let say $y=fg\left(x\right)$ , where $f\,$ and $g\,$ are basic functions which we know how to differentiate. Since the $f\,$ is "applied" to $g\left(x\right)$ , it might be easier to us to rewrite it as $f\left(u\right)$ , where $u=g\left(x\right).$

So how would we differentiate it?

$\frac{dy}{dx}$ $=\frac{d\left[fg\left(x\right)\right]}{dx}$ $=\frac{d\left[f\left(u\right)\right]}{dx}$

Now, we can differentiate $\frac{d\left[f\left({\color{Blue}x}\right)\right]}{d{\color{Blue}x}}$ directly, but not $\frac{d\left[f\left({\color{Red}u}\right)\right]}{d{\color{Blue}x}}$ .

For example
In other words,

So let's look back at $\frac{d\left[f\left(u\right)\right]}{dx}$

We know that we can differentiate $f\left(u\right)$
But surely we CAN'T just change $\frac{d\left[f\left(u\right)\right]}{dx}$ to $\frac{d\left[f\left(u\right)\right]}{du}$ by writing $\frac{d\left[f\left(u\right)\right]}{{\color{Blue}dx}}=\frac{d\left[f\left(u\right)\right]}{{\color{Red}du}}$
We need to
Thus, $\frac{d\left[f\left(u\right)\right]}{{\color{Blue}dx}}=\frac{d\left[f\left(u\right)\right]}{{\color{Red}du}}\times$ $\frac{{\color{Red}du}}{{\color{Blue}dx}}$

Once again, $\frac{d\left[fg\left(x\right)\right]}{dx}$ $=\frac{d\left[f\left({\color{Red}u}\right)\right]}{dx}$ $=\frac{d\left[f\left({\color{Red}u}\right)\right]}{{\color{Red}du}}\times\frac{{\color{Red}du}}{dx}$

[f(x)]^n

M) $\frac{d}{dx}{\left[f\left(x\right)\right]}^{n}$ $=n{\left[f\left(x\right)\right]}^{n-1}\cdot f'\left(x\right)$

Remember :

Prove

Analyze :
When we see , we see
- Let $f\left(x\right)=u$
- $\frac{d}{dx}{\left[f\left(x\right)\right]}^{n}=\frac{d}{dx}\left({u}^{n}\right)$
  $=\frac{d}{du}\left({u}^{n}\right)\times$ $\frac{du}{dx}$
  $=n{u}^{n-1}\left(\frac{du}{dx}\right)$
  $=n{\left[f\left(x\right)\right]}^{n-1}\cdot f'\left(x\right)$

Examples

Find $f'\left(x\right)$ for the following

$f'\left(x\right)=18{\left(3x-7\right)}^{5}$
- Analyze : Something power 6 . Note: that something is $\left(3x-7\right)$ and NOT ${\left(3x-7\right)}^{6}$
  - First, we write $f'\left(x\right)$
  - Bring down the index $f'\left(x\right)={\color{Red}6}$
  - Copy exactly that something $f'\left(x\right)=6{\color{Red}\left(3x-7\right)}$
  - Minus one to the index $f'\left(x\right)=6{\left(3x-7\right)}^{{\color{Red}5}}$
  - Simplify $=18{\left(3x-7\right)}^{5}$

$f'\left(x\right)=-\frac{12}{{\left(4x+1\right)}^{2}}$
- Analyze : Constant over something, which means it is something power $-1\,$
  - $f\left(x\right)=\frac{3}{4x+1}$ $=3{\left(4x+1\right)}^{-1}$
  - $f'\left(x\right)={\color{Red}-3}$
  - $f'\left(x\right)=-3{\color{Red}\left(4x+1\right)}$
  - $f'\left(x\right)=-3{\left(4x+1\right)}^{{\color{Red}-2}}$
  - $f'\left(x\right)=-3{\left(4x+1\right)}^{-2}{\color{Red}\cdot4}$
  - $=-\frac{12}{{\left(4x+1\right)}^{2}}$
  - Alternatively
    $f'\left(x\right)=\frac{{\color{Red}-3}}{}$
    $f'\left(x\right)=\frac{-3}{{\color{Red}\left(4x+1\right)}}$
    $f'\left(x\right)=\frac{-3}{{\left(4x+1\right)}^{{\color{Red}2}}}$
    $f'\left(x\right)=\frac{-3}{{\left(4x+1\right)}^{2}}{\color{Red}\cdot4}$
    $=-\frac{12}{{\left(4x+1\right)}^{2}}$

- $f'\left(x\right)=3{\left({x}^{2}+\frac{4}{x}+1 \right)}^{2}\left(2x-\frac{4}{{x}^{2}} \right)$
- Note:
  There are some common mistakes when the something gets more complicated like above

- Since we know that $=\frac{1}{2\sqrt{x}}$ , we can actually do it directly
  - $f'\left(x\right)=\frac{3}{2\sqrt{3x+1}}$

- - $\begin{align} f\left(x\right) &=\frac{1}\sqrt{1-{x}^{2}}\\ &={\left(1-{x}^{2}\right)}^{-\frac{1}{2}}\\ f'\left(x\right) &=-\frac{1}{2}{\left(1-{x}^{2}\right)}^{-\frac{3}{2}}\left(-2x \right)\\ &=\frac{x}{{\left(1-{x}^{2}\right)}^{\frac{3}{2}}} \end{align}$

$f'\left(x\right)={\left(x+1\right)}^{2}{\left(2x-1\right)}^{3}\left(14x+5 \right)$
- Analyze :
  - Copy first thing $f'\left(x\right)={\color{Red}{\left(x+1\right)}^{3}}$
  - Plus copy second thing $f'\left(x\right)={\left(x+1\right)}^{3}\cdot 4{\left(2x-1\right)}^{3}\left(2 \right)+{\color{Red}{\left(2x-1\right)}^{4}}$
  - Simplify by

$f'\left(x\right)=\frac{2{\left(4x-1 \right)}^{2}\left(2x+13\right)}{{\left(x+2 \right)}^{3}}$
- Analyze : Something over something
  - Differentiating $f'\left(x\right)=\frac{{\left(x+2 \right)}^{2}\cdot3{\left(4x-1 \right)}^{2}\left(4\right)-{\left(4x-1\right)}^{3}\cdot2\left(x+2\right)\left(1\right)}{{\left(x+2 \right)}^{4}}$
  - Factorizing $f'\left(x\right)=\frac{{\color{Red}2\left(x+2 \right){\left(4x-1 \right)}^{2}}\qquad\qquad\qquad}{{\left(x+2 \right)}^{4}}$
  - $f'\left(x\right)=\frac{2\left(x+2 \right){\left(4x-1 \right)}^{2}{\color{Red}\left[6\left(x+2\right)-\left(4x-1\right)\right]}}{{\left(x+2 \right)}^{4}}$

Exercise 3

Find $f'\left(x\right)$ for the following

a) $f\left(x\right)={\left(1-5x\right)}^{7}-{\left({x}^{2}-1\right)}^{-3}$

$f'\left(x\right)=-35{\left(1-5x\right)}^{6}+6x{\left({x}^{2}-1\right)}^{-4}$
- $\begin{align}f'\left(x\right)&=7{\left(1-5x\right)}^{6}\left(-5\right)-3{\left({x}^{2}-1\right)}^{-4}\left(2x\right)\\ &=-35{\left(1-5x\right)}^{6}+6x{\left({x}^{2}-1\right)}^{-4} \end{align}$

b) $f\left(x\right)=\frac{5}{7x-1}+\frac{3}{{\left(x+1\right)}^{5}}$

$f'\left(x\right)=-\frac{35}{{\left(7x-1\right)}^{2}}-\frac{15}{{\left(x+1\right)}^{6}}$
- $\begin{align} f\left(x\right) & =\frac{5}{7x-1}+\frac{3}{{\left(x+1\right)}^{5}}\\ & =5{\left(7x-1\right)}^{-1}+3{\left(x+1\right)}^{-5} \\ f'\left(x\right)&=-5{\left(7x-1\right)}^{-2}\left(7\right)-15{\left(x+1\right)}^{-6} \\ & =-\frac{35}{{\left(7x-1\right)}^{2}}-\frac{15}{{\left(x+1\right)}^{6}} \end{align}$

c) $f\left(x\right)={\left({x}^{2}-2\sqrt{x}+3 \right)}^{4}$

$f'\left(x\right)=4{\left({x}^{2}-2\sqrt{x}+3 \right)}^{3}\left(2x-\frac{1}{\sqrt{x}}\right)$
- $\begin{align} f'\left(x\right)& =4{\left({x}^{2}-2\sqrt{x}+3 \right)}^{3}\left(2x-\frac{2}{2\sqrt{x}}\right)\\ & =4{\left({x}^{2}-2\sqrt{x}+3 \right)}^{3}\left(2x-\frac{1}{\sqrt{x}}\right)\\ \end{align}$

d) $f\left(x\right)=\sqrt{{x}^{2}-4}+\frac{2}{\sqrt{x+3}}$

$f'\left(x\right)=\frac{x}{\sqrt{{x}^{2}-4}}-\frac{1}{{\left(x+3 \right)}^{\frac{3}{2}}}$
- $\begin{align} f\left(x\right)&=\sqrt{{x}^{2}-4}+\frac{2}{\sqrt{x+3}}\\ &=\sqrt{{x}^{2}-4}+2{\left(x+3\right)}^{-\frac{1}{2}}\\ f'\left(x\right)&=\frac{2x}{2\sqrt{{x}^{2}-4}}-{\left(x+3 \right)}^{-\frac{3}{2}}\\ &=\frac{x}{\sqrt{{x}^{2}-4}}-\frac{1}{{\left(x+3 \right)}^{\frac{3}{2}}} \end{align}$

e) $f\left(x\right)=\frac{1}{{x}^{2}+2x+8}$

$f'\left(x\right)=-\frac{2\left(x+1\right)}{{\left({x}^{2}+2x+8\right)}^{2}}$
- $\begin{align} f'\left(x\right)& =-\frac{2x+2}{{\left({x}^{2}+2x+8\right)}^{2}}\\ & =-\frac{2\left(x+1\right)}{{\left({x}^{2}+2x+8\right)}^{2}}\\ \end{align}$

f) $f\left(x\right)={\left(2x+5\right)}^{6}{\left(3x+2\right)}^{3}$

$f'\left(x\right)=3{\left(2x+5\right)}^{5}{\left(3x+2\right)}^{2}\left(18x+23\right)$
- $\begin{align} f\left(x\right)& ={\left(2x+5\right)}^{6}{\left(3x+2\right)}^{3}\\ f'\left(x\right)& ={\left(2x+5\right)}^{6}\cdot 3{\left(3x+2\right)}^{2}\left(3\right)+{\left(3x+2\right)}^{3}\cdot6{\left(2x+5\right)}^{5}\left(2\right)\\ &=3{\left(2x+5\right)}^{5}{\left(3x+2\right)}^{2}\left[3\left(2x+5\right)+4\left(3x+2\right)\right]\\ &=3{\left(2x+5\right)}^{5}{\left(3x+2\right)}^{2}\left(18x+23\right) \end{align}$

g) $f\left(x\right)={\left(x+1\right)}^{4}{\left({x}^{2}-7\right)}^{2}$

$f'\left(x\right)=4{\left(x+1\right)}^{3}\left({x}^{2}-7\right)\left(2{x}^{2}+x-7\right)$
- $\begin{align} f\left(x\right)&={\left(x+1\right)}^{4}{\left({x}^{2}-7\right)}^{2}\\ f'\left(x\right)& ={\left(x+1\right)}^{4}\cdot2\left({x}^{2}-7\right)\left(2x\right)+{\left({x}^{2}-7\right)}^{2}\cdot4{\left(x+1\right)}^{3}\\ &=4{\left(x+1\right)}^{3}\left({x}^{2}-7\right)\left[x\left(x+1\right)+\left({x}^{2}-7\right)\right]\\ &=4{\left(x+1\right)}^{3}\left({x}^{2}-7\right)\left(2{x}^{2}+x-7\right) \end{align}$

h) $f\left(x\right)=\frac{{\left(2x-5\right)}^{2}}{{\left(1-x\right)}^{3}}$

$f'\left(x\right)= \frac{\left(2x-5\right)\left(2x-11\right)}{{\left(1-x\right)}^{4}}$
- $\begin{align} f\left(x\right)&=\frac{{\left(2x-5\right)}^{2}}{{\left(1-x\right)}^{3}}\\ f'\left(x\right)& =\frac{{\left(1-x\right)}^{3}\cdot2\left(2x-5\right)\left(2\right)-{\left(2x-5\right)}^{2}\cdot3{\left(1-x\right)}^{2}\left(-1 \right)}{{\left(1-x\right)}^{6}}\\ & =\frac{{\left(1-x\right)}^{2}\left(2x-5\right)\left[4\left(1-x\right)+3\left(2x-5\right)\right]}{{\left(1-x\right)}^{6}}\\ &=\frac{\left(2x-5\right)\left(2x-11\right)}{{\left(1-x\right)}^{4}} \end{align}$

e^[f(x)]

N) $\frac{d}{dx}{e}^{f\left(x\right)}$ $={e}^{f\left(x\right)}\cdot f'\left(x\right)$ $=f'\left(x\right){e}^{f\left(x\right)}$

Remember :
Note :
Important :
Note: The ${\color{Blue}f'\left(x\right)}{e}^{{\color{Blue}f\left(x\right)}}$ has nothing to do with $\frac{d}{dx}x^{n}={\color{Blue}n}x^{{\color{Blue}n-1}}$ .

You can try the prove yourself, but it is more important for us to remember and use the formula correctly for now.

Comparing with the basic formula

$\frac{d}{dx}{e}^{x}$ $={e}^{x} \,$
$\frac{d}{dx}{e}^{f\left(x\right)}$ $={e}^{f\left(x\right)}\cdot f'\left(x\right)$

Now, for
- Can you guess the formula for $\frac{d}{dx}{a}^{f\left(x\right)}$ ?
- $\frac{d}{dx}{a}^{f\left(x\right)}$

O) $\frac{d}{dx}{a}^{f\left(x\right)}$ $=f'\left(x\right){a}^{f\left(x\right)}\ln a$

Examples

Differentiate the following w.r.t. $x\,$

$\frac{dy}{dx}=2{e}^{2x}$
- Analyze : $e\,$ power something.
$y={e}^{3x+1}\,$ $\frac{dy}{dx}=3{e}^{3x+1}$
$y={e}^{1-{x}^{2}}\,$ $\frac{dy}{dx}=-2x{e}^{1-{x}^{2}}$
$y={e}^{-x}\,$ $\frac{dy}{dx}=-{e}^{-x}$
$\frac{dy}{dx}=-3{e}^{-3x}$ or $\frac{dy}{dx}=-\frac{3}{{e}^{3x}}$
- Analyze :
  - $y=\frac{1}{{e}^{3x}}$ $={e}^{-3x}\,$
  - $\frac{dy}{dx}=-3{e}^{-3x}$ or $\frac{dy}{dx}=-\frac{3}{{e}^{3x}}$
$\frac{dy}{dx}=\frac{1}{2}{e}^{\frac{1}{2}x}$ or $\frac{dy}{dx}=\frac{1}{2}\sqrt{{e}^{x}}$
- Analyze :
  - $y=\sqrt{{e}^{x}}$ $={e}^{\frac{1}{2}x}$
  - $\frac{dy}{dx}=\frac{1}{2}{e}^{\frac{1}{2}x}$ or $\frac{dy}{dx}=\frac{1}{2}\sqrt{{e}^{x}}$
$\frac{dy}{dx}={e}^{\sin x}\cos x \,$
$y={e}^{\tan x}\,$ $\frac{dy}{dx}={e}^{\tan x}{\sec}^{2}x \,$
$\frac{dy}{dx}={e}^{x}+3{e}^{-3x}\,$
- Analyze :
  - $y=\frac{{e}^{2x}-{e}^{-2x}}{{e}^{x}}$ $={e}^{x}-{e}^{-3x}\,$
  - $\frac{dy}{dx}={e}^{x}+3{e}^{-3x}$

ln[f(x)]

P) $\frac{d}{dx}\ln\left[f\left(x \right)\right]$ $=\frac{1}{f\left(x\right)}\cdot f'\left(x\right)$ $=\frac{f'\left(x\right)}{f\left(x\right)}$

Remember :
Note :

Comparing with the other composite formula, we see that the pattern is really the same:

- $\frac{d}{dx}{\left[{\color{Blue}f\left(x\right)}\right]}^{n} =n{\left[{\color{Blue}f\left(x\right)}\right]}^{n-1}\cdot {\color{Red}f'\left(x\right)}$
- $\frac{d}{dx}{e}^{{\color{Blue}f\left(x\right)}}={\color{Red}f'\left(x\right)}{e}^{{\color{Blue}f\left(x\right)}}$
- $\frac{d}{dx}\ln\left[{\color{Blue}f\left(x\right)}\right]=\frac{{\color{Red}f'\left(x\right)}}{{\color{Blue}f\left(x\right)}}$

Whereby we just

substitute any ${\color{Blue}x}$ with ${\color{Blue}f\left(x\right)}$
and multiply an extra ${\color{Red}f'\left(x\right)}$

Using Index Laws

Note that

$\ln xy\,$ $=\ln x+\ln y\,$
$\ln \frac{x}{y}$ $=\ln x-\ln y\,$
$\ln\left(x+y\right)$ Nothing we can do. ;-)
$\ln {x}^{n}\,$ $=n\ln x\,$

Take a good look at each law. Which side (right or left) would be easier to differentiate? The right hand side, of course.

Thus, ALWAYS check whether we can simplify first when we differentiate $ln$ functions.

Examples

$\frac{dy}{dx}=\frac{3}{3x+1}$
- Analyze : $\ln\,$ something.
  - Copy that something, put it below $\frac{dy}{dx}=\frac{}{{\color{Red}3x+1}}$
$y=\ln\left({x}^2+1\right)$ $\frac{dy}{dx}=\frac{2x}{{x}^2+1}$
$y=\ln \left(\sin x\right)$ $\frac{dy}{dx}=\frac{\cos x}{\sin x}$ or $\frac{dy}{dx}=\cot x$
$\frac{dy}{dx}=\frac{3}{x+1}$
- Analyze : $\ln\,$ something. Can we simpflify? Yes.
  - $y=\ln{\left(x+1\right)}^{3}$ $=3\ln\left(x+1\right)$
  - $\frac{dy}{dx}=\frac{3}{x+1}$

$\frac{dy}{dx}=\frac{1}{2x}$
- $=\frac{1}{2}\ln x$
  - $\frac{dy}{dx}=\frac{1}{2x}$

$\frac{dy}{dx}=-\frac{4}{x}$
- $=\ln 5-4 \ln x\,$ ( Note that $\ln 5\,$ is a constant.)
  - $\frac{dy}{dx}=-\frac{4}{x}$

$\frac{dy}{dx}=\frac{3}{2\left(3x-1\right)}$
- $=\frac{1}{2}\ln\left(3x-1\right)$
  - $\frac{dy}{dx}=\frac{3}{2\left(3x-1\right)}$

$\frac{dy}{dx}=-\frac{1}{3\left(1-x\right)}-\frac{2}{3\left(1+x\right)}$
- $=\frac{1}{3}\ln\left(1-x\right)-\frac{2}{3}\ln\left(1+x\right)$
  - $\frac{dy}{dx}=-\frac{1}{3\left(1-x\right)}-\frac{2}{3\left(1+x\right)}$

$\frac{dy}{dx}=\frac{2}{x}+1$
- $=2\ln x+\ln {e}^{x}\,$ $=2\ln x+x\,$
  - $\frac{dy}{dx}=\frac{2}{x}+1$

sin[f(x)], cos[f(x)], tan[f(x)]

You should be able to guess the following formulas already

Q) $\frac{d}{dx}\sin\left[f\left(x\right)\right]$ $=f'\left(x\right)\cos\left[f\left(x\right)\right]$

R) $\frac{d}{dx}\cos\left[f\left(x\right)\right]$ $=-f'\left(x\right)\sin\left[f\left(x\right)\right]$

S) $\frac{d}{dx}\tan\left[f\left(x\right)\right]$ $=f'\left(x\right){\sec}^{2}\left[f\left(x\right)\right]$

If you still can't see why
- $\frac{d}{dx}\sin{\color{Blue}x}=\cos{\color{Blue}x}$
- $\frac{d}{dx}\sin{\color{Blue}f\left(x\right)}={\color{Red}f'\left(x\right)}\cos{\color{Blue}f\left(x\right)}$
- $\frac{d}{dx}\cos{\color{Blue}x}=-\sin{\color{Blue}x}$
- $\frac{d}{dx}\cos{\color{Blue}f\left(x\right)}=-{\color{Red}f'\left(x\right)}\sin{\color{Blue}f\left(x\right)}$
- $\frac{d}{dx}\tan{\color{Blue}x}={\sec}^{2}{\color{Blue}x}$
- $\frac{d}{dx}\tan{\color{Blue}f\left(x\right)}=-{\color{Red}f'\left(x\right)}{\sec}^{2}{\color{Blue}f\left(x\right)}$

Remember :

Note :

Common Mistakes

Remember that, for sin something, for example, we need to

change it to cos
copy that something
then only multiply the derivative

It is very easy to carelessly skip the second step, so take the time to learn the structure correctly. For example, when we do something such as $y=\sin \left(\ln x\right)$ , if we do it too quickly, we might carelessly end up with $\frac{dy}{dx}=\cos \left(\frac{1}{x}\right)$ when it should have been $\frac{dy}{dx}=\frac{1}{x}\cos \left(\ln x\right)$ .

Examples

$=3\cos 3x\,$
- Analysis: sin something
  - copy that something $\frac{d}{dx}\sin 3x=\qquad\cos{\color{Red}3x}$
$\frac{d}{dx}\tan \left(3x-1\right)$ $=3{\sec}^{2}\left(3x-1\right)\,$
$=-2x\sin\left({x}^2+1\right)\,$
- The more complicated it gets, the more we need to make sure our basic steps are correct

$=-\frac{1}{{x}^{2}}\cos\left(\frac{1}{x}\right)$
- $={\color{Red}\qquad \cos \left(\frac{1}{x}\right)}$
  - $\frac{d}{dx}\sin \left(\frac{1}{x}\right)$ $={\color{Red}-\frac{1}{{x}^{2}}}\cos \left(\frac{1}{x}\right)$
  - Note:

$\frac{d}{dx}\cos \sqrt{x}$ $=-\frac{1}{2\sqrt{x}}\sin\sqrt{x}$
$\frac{d}{dx}\tan \left({e}^{x}\right)$ $={e}^{x}{\sec}^{2}\left({e}^{x}\right)$

Exercise 4

Differentiate the following w.r.t. $x\,$

a) $y={e}^{3x}-4{e}^{-2x}+{e}^{{x}^{2}}+3\sqrt{{e}^{x}}+\frac{1}{\sqrt{{e}^{x}}}$

$\frac{dy}{dx}=3{e}^{3x}+8{e}^{-2x}+2x{e}^{{x}^{2}}+\frac{3}{2}\sqrt{{e}^{x}}-\frac{1}{2\sqrt{{e}^{x}}}$
- $\begin{align} y&={e}^{3x}-4{e}^{-2x}+{e}^{{x}^{2}}+3\sqrt{{e}^{x}}+\frac{1}{\sqrt{{e}^{x}}}\\ &={e}^{3x}-4{e}^{-2x}+{e}^{{x}^{2}}+3{e}^{\frac{1}{2}x}+{e}^{-\frac{1}{2}x}\\ \frac{dy}{dx}&=3{e}^{3x}+8{e}^{-2x}+2x{e}^{{x}^{2}}+\frac{3}{2}{e}^{\frac{1}{2}x}-\frac{1}{2}{e}^{-\frac{1}{2}x}\\ &=3{e}^{3x}+8{e}^{-2x}+2x{e}^{{x}^{2}}+\frac{3}{2}\sqrt{{e}^{x}}-\frac{1}{2\sqrt{{e}^{x}}}\\ \end{align}$

b) $y={e}^{\cos x}-2{e}^{\frac{3}{x}}+4{e}^{\sqrt{x}+1}+{e}^{\tan x+\ln x}$

$\frac{dy}{dx}=-{e}^{\cos x}\sin x+\frac{6}{{x}^{2}}{e}^{\frac{3}{x}}+\frac{2}{\sqrt{x}}{e}^{\sqrt{x}+1}+\left({\sec}^{2}x+\frac{1}{x}\right){e}^{\tan x+\ln x}$
- $\begin{align} \frac{dy}{dx}&=\left(-\sin x\right){e}^{\cos x}-2\left(-\frac{3}{{x}^{2}}\right){e}^{\frac{3}{x}}+4\left(\frac{1}{2\sqrt{x}}\right){e}^{\sqrt{x}+1}+\left({\sec}^{2}x+\frac{1}{x}\right){e}^{\tan x+\ln x}\\ &=-{e}^{\cos x}\sin x+\frac{6}{{x}^{2}}{e}^{\frac{3}{x}}+\frac{2}{\sqrt{x}}{e}^{\sqrt{x}+1}+\left({\sec}^{2}x+\frac{1}{x}\right){e}^{\tan x+\ln x}\\ \end{align}$

c) $y=\frac{4{e}^{3x}-2{e}^{-3x}}{{e}^{2x}}$

$\frac{dy}{dx}=4{e}^{x}+10{e}^{-5x}$
- $\begin{align} y& =\frac{4{e}^{3x}-2{e}^{-3x}}{{e}^{2x}}\\ &=4{e}^{x}-2{e}^{-5x}\\ \frac{dy}{dx}& =4{e}^{x}+10{e}^{-5x} \end{align}$

d) $y=\ln\left(2x-1\right)-2\ln\left(7-3x \right)$ $\frac{dy}{dx}= \frac{2}{2x-1}+\frac{6}{7-3x}$

- $\begin{align} \frac{dy}{dx} &= \frac{2}{2x-1}-2\left(\frac{-3}{7-3x}\right)\\ &= \frac{2}{2x-1}+\frac{6}{7-3x} \\ \end{align}$

e) $y=\ln\left(3{x}^{2}-4\right)$

$\frac{dy}{dx}= \frac{6x}{3{x}^{2}-4}$

f) $y=3\ln\left(\cos x\right)-2\ln\left(\sin x \right)$

$\frac{dy}{dx}= -\frac{3\sin x}{\cos x}-\frac{2\cos x}{\sin x}$

g) $y=\ln\left({e}^{x}+4\right)+\ln\left(\sin x +\cos x \right)$

$\frac{dy}{dx}= \frac{{e}^{x}}{{e}^{x}+4}+\frac{\cos x-\sin x}{\sin x+\cos x}$

h) $y=3\ln\left(\frac{5}{{x}^{2}}\right)-\ln\sqrt{\frac{4}{{x}^{3}}}$

$\frac{dy}{dx}=-\frac{9}{2x}$
- $\begin{align} y& =3\ln\left(\frac{5}{{x}^{2}}\right)-\ln\sqrt{\frac{4}{{x}^{3}}}\\ & =3\ln 5 -6\ln x -\left( \frac{1}{2}\ln 4 -\frac{3}{2}\ln x\right)\\ \frac{dy}{dx} &= -\frac{6}{x}+\frac{3}{2x}\\ & =-\frac{9}{2x} \end{align}$

i) $y=\ln\sqrt{1-{x}^{2}}+\ln {\left(3{x}^{2}+4 \right)}^{3}$

$\frac{dy}{dx}=-\frac{x}{1-{x}^{2}}+\frac{18x}{3{x}^{2}+4}$
- $\begin{align} y& =\ln\sqrt{1-{x}^{2}}+\ln {\left(3{x}^{2}+4 \right)}^{3} \\ & =\frac{1}{2}\ln\left(1-{x}^{2}\right)+3\ln\left(3{x}^{2}+4 \right) \\ \frac{dy}{dx} &= \frac{-2x}{2\left(1-{x}^{2}\right)}+\frac{3\left(6x\right)}{3{x}^{2}+4}\\ & =-\frac{x}{1-{x}^{2}}+\frac{18x}{3{x}^{2}+4} \end{align}$

j) $y=\ln\left(\frac{1+x}{1-x}\right)$

$\frac{dy}{dx}=\frac{2}{\left(1+x\right)\left(1-x\right)}$
- $\begin{align} y& =\ln\left(\frac{1+x}{1-x}\right)\\ & =\ln\left(1+x\right)-\ln\left(1-x\right) \\ \frac{dy}{dx} &=\frac{1}{1+x}-\frac{\left(-1 \right)}{1-x} \\ & =\frac{1-x+1+x}{\left(1+x\right)\left(1-x\right)}\\ & =\frac{2}{\left(1+x\right)\left(1-x\right)} \end{align}$

k) $y=\ln\left[{\left(2x+1\right)}^{3}{\left(x-2\right)}^{5}\right]$

$\frac{dy}{dx}=\frac{6}{2x+1}+\frac{5}{x-2}$
- $\begin{align} y& =\ln\left[{\left(2x+1\right)}^{3}{\left(x-2\right)}^{5}\right]\\ & =3\ln\left(2x+1\right)+5\ln\left(x-2\right)\\ \frac{dy}{dx}& =\frac{6}{2x+1}+\frac{5}{x-2}\\ \end{align}$

l) $y=\ln\frac{{\left({x}^{2}-3\right)}^{4}}{\left(1-2x \right)}$

$\frac{dy}{dx}=\frac{8x}{{x}^{2}-3}+\frac{2}{1-2x}$
- $\begin{align} y&=\ln\frac{{\left({x}^{2}-3\right)}^{4}}{\left(1-2x \right)}\\ &=4\ln\left({x}^{2}-3\right)-\ln\left(1-2x \right)\\ \frac{dy}{dx}&=\frac{8x}{{x}^{2}-3}+\frac{2}{1-2x} \end{align}$

m) $y=\ln\sqrt{\frac{1+x}{1-x}}$

$\frac{dy}{dx}=\frac{1}{\left(1+x\right)\left(1-x\right)}$
- $\begin{align} y& =\ln\sqrt{\frac{1+x}{1-x}}\\ & =\frac{1}{2}\left[\ln\left(1+x\right)-\ln\left(1-x\right)\right] \\ \frac{dy}{dx} &=\frac{1}{2}\left[ \frac{1}{1+x}-\frac{\left(-1 \right)}{1-x}\right] \\ & =\frac{1}{2}\left[ \frac{1-x+1+x}{\left(1+x\right)\left(1-x\right)}\right]\\ & =\frac{1}{\left(1+x\right)\left(1-x\right)} \end{align}$

n) $y=\ln\sqrt[3]{\frac{{\left(3x-1\right)}^{4}}{{x}^{2}}}$

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

o) $y=\sin3x+3\cos\left(2x-1\right)+\tan\left(1-x\right)$

$\frac{dy}{dx}=3\cos 3x-6\sin\left(2x-1\right)-{\sec}^{2}\left(1-x\right)$

p) $y=\cos\left(\pi x\right)+\sin\left(\frac{\pi}{3}x\right)-\sin\left(\pi+x\right)$

$\frac{dy}{dx}=-\pi\sin\left(\pi x\right)+\frac{\pi}{3}\cos\left(\frac{\pi}{3}x\right)-\cos\left(\pi+x\right)$

q) $y=\tan\left({e}^{x}+2\right)+\cos\left(2\sqrt{x}\right)-\sin\left(\frac{3}{x}\right)$

$\frac{dy}{dx}={e}^{x}{\sec}^{2}\left({e}^{x}+2\right)-\frac{\sin\left(2\sqrt{x}\right)}{\sqrt{x}}+\frac{3}{{x}^{2}}\cos\left(\frac{3}{x}\right)$
- $\begin{align} \frac{dy}{dx}&={e}^{x}{\sec}^{2}\left({e}^{x}+2\right)-\frac{2}{2\sqrt{x}}\sin\left(2\sqrt{x}\right)-\left( -\frac{3}{{x}^{2}}\right)\cos\left(\frac{3}{x}\right)\\ &={e}^{x}{\sec}^{2}\left({e}^{x}+2\right)-\frac{\sin\left(2\sqrt{x}\right)}{\sqrt{x}}+\frac{3}{{x}^{2}}\cos\left(\frac{3}{x}\right) \end{align}$

r) $y=\sqrt{{e}^{x}}+{e}^{\sqrt{x}}+\frac{1}{{e}^{x}}+{e}^{\frac{1}{x}}$

$\frac{dy}{dx}=\frac{1}{2}\sqrt{{e}^{x}}+\frac{{e}^{\sqrt{x}}}{2\sqrt{x}}-{e}^{-x}-\frac{1}{{x}^{2}}{e}^{\frac{1}{x}}$
- $\begin{align} y&=\sqrt{{e}^{x}}+{e}^{\sqrt{x}}+\frac{1}{{e}^{x}}+{e}^{\frac{1}{x}}\\ &={e}^{\frac{1}{2}x}+{e}^{\sqrt{x}}+{e}^{-x}+{e}^{\frac{1}{x}}\\ \frac{dy}{dx}&=\frac{1}{2}{e}^{\frac{1}{2}x}+\frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}-{e}^{-x}+\left( -\frac{1}{{x}^{2}}\right){e}^{\frac{1}{x}}\\ &=\frac{1}{2}\sqrt{{e}^{x}}+\frac{{e}^{\sqrt{x}}}{2\sqrt{x}}-{e}^{-x}-\frac{1}{{x}^{2}}{e}^{\frac{1}{x}} \end{align}$

s) $y=\sin\left({e}^{x}\right)+{e}^{\sin x}+\ln\left(\sin x\right)+\sin\left(\ln x\right)$

$\frac{dy}{dx}={e}^{x}\cos\left({e}^{x}\right)+{e}^{\sin x}\cos x+\frac{\cos x}{\sin x}+\frac{1}{x}\cos\left(\ln x\right)$

Differentiation Part2

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Chain Rule

Composite Functions

[f(x)]^n

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Exercise 3

e^[f(x)]

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ln[f(x)]

Using Index Laws

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sin[f(x)], cos[f(x)], tan[f(x)]

Common Mistakes

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Exercise 4

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