Differentiation Part3

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More [f(x)]^n

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

Note that $f(x)\,$ can refer to ANY functions/combination of functions. Here we will look at slightly more complicated functions (compared to polynomials/rational functions earlier on)

Common Mistakes: It is very common that as we move to more complicated functions, students tend to skip a step or two as they get too distracted with differentiating the something. Always remember however complicated is that something, the main structure of the derivative never changes, and the derivative of that something is always the last thing we do, after we settle the main structure. I can't stress enough the importance of checking it once we are done. It takes maybe an 10% of the time you would have used to solve the question, but can get rid of most careless mistakes.

Examples

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

$\frac{dy}{dx}=4{e}^{x}{\left({e}^{x}+3\right)}^{3}$
- Lets revise the steps again even though this would be relatively simple
  - Something power n
  - Bring down index $\frac{dy}{dx}={\color{Red}4}$
  - Copy exactly that something $\frac{dy}{dx}=4{\color{Red}\left({e}^{x}+3\right)}$
  - Minus one to the index $\frac{dy}{dx}=4{\left({e}^{x}+3\right)}^{{\color{Red}3}}$

$\frac{dy}{dx}=-\frac{2}{x{\left(\ln x+2\right)}^{3}}$
- Something power n
  - $y=\frac{1}{{\left( \ln x+2\right)}^{2}}$ $={\left( \ln x+2\right)}^{-2}$
  - $\frac{dy}{dx}={\color{Red}-2}$
  - $\frac{dy}{dx}=-2{\color{Red}\left( \ln x+2\right)}$
  - $\frac{dy}{dx}=-2{\left( \ln x+2\right)}^{{\color{Red}-3}}$
  - $\frac{dy}{dx}=-2{\left( \ln x+2\right)}^{-3}{\color{Red}\left(\frac{1}{x}\right)}$

$\frac{dy}{dx}=\frac{\cos x}{\sqrt{1+2\sin x}}$
- Something power n
  - $y=\sqrt{1+2\sin x}$ $={\left(1+2\sin x\right)}^{\frac{1}{2}}$
  - $\frac{dy}{dx}={\color{Red}\frac{1}{2}}$
  - $\frac{dy}{dx}=\frac{1}{2}{\color{Red}\left(1+2\sin x\right)}$
  - $\frac{dy}{dx}=\frac{1}{2}{\left(1+2\sin x\right)}^{{\color{Red}-\frac{1}{2}}}$
  - $\frac{dy}{dx}=\frac{1}{2}{\left(1+2\sin x\right)}^{-\frac{1}{2}}{\color{Red}\left(2\cos x\right)}$

$\frac{dy}{dx}=\frac{\sin x-\cos x}{{\left(\sin x+\cos x\right)}^{2}}$
- 1 over something, which means something power n
  - $y=\frac{1}{\sin x+\cos x}$ $={\left(\sin x+\cos x\right)}^{-1}$
  - $\frac{dy}{dx}={\color{Red}-}$
  - $\frac{dy}{dx}=-{\color{Red}\left(\sin x+\cos x\right)}$
  - $\frac{dy}{dx}=-{\left(\sin x+\cos x\right)}^{{\color{Red}-2}}$
  - $\frac{dy}{dx}=-{\left(\sin x+\cos x\right)}^{-2}{\color{Red}\left(\cos x-\sin x\right)}$
  - Alternatively
    $y=\frac{1}{\sin x+\cos x}$
    $\frac{dy}{dx}={\color{Red}-}$
    $\frac{dy}{dx}=-\frac{\quad}{{\color{Red}\left(\sin x+\cos x\right)}}$
    $\frac{dy}{dx}=-\frac{\quad}{{\left(\sin x+\cos x\right)}^{{\color{Red}2}}}$
    $\frac{dy}{dx}=-\frac{{\color{Red}\cos x- \sin x}}{{\left(\sin x+\cos x\right)}^{2}}$

$\frac{dy}{dx}=-\frac{2\left(6{e}^{2x}+{e}^{-x}\right)}{{\left(3{e}^{2x}-{e}^{-x}\right)}^{\frac{3}{2}}}$
- something power n
  - $y=\frac{4}{\sqrt{3{e}^{2x}-{e}^{-x}}}$ $=4{\left(3{e}^{2x}-{e}^{-x}\right)}^{-\frac{1}{2}}$
  - $\frac{dy}{dx}={\color{Red}-2}$
  - $\frac{dy}{dx}=-2{\color{Red}\left(3{e}^{2x}-{e}^{-x}\right)}$
  - $\frac{dy}{dx}=-2{\left(3{e}^{2x}-{e}^{-x}\right)}^{{\color{Red}-\frac{3}{2}}}$
  - $\frac{dy}{dx}=-2{\left(3{e}^{2x}-{e}^{-x}\right)}^{-\frac{3}{2}}{\color{Red}\left(6{e}^{2x}+{e}^{-x}\right)}$

sin^n x, cos^n x, tan^n x

The most important thing here isn't memorizing any new formulas, but understanding some notation

${\sin}^{n}x\,$ $={\left(\sin x\right)}^{n}$
${\cos}^{n}x\,$ $={\left(\cos x\right)}^{n}$
Note: $n\,$ must be positive integer

Simply said, those aren't even formulas, it simply is a shorthand, a shortcut so that we don't need to write the brackets. Understand this and basically you are set.

More examples

${\sin}^{2}x\,$ $={\left(\sin x\right)}^{2}$
${\cos}^{5}x\,$ $={\left(\cos x\right)}^{5}$
${\tan}^{3}x\,$ $={\left(\tan x\right)}^{3}$
Note:
- $\frac{1}{\sin x}$ CANNOT be written as ${\sin}^{-1}x\,$
- $\sqrt{\sin x}$ also wont be written in the form of ${\sin}^{n}x\,$

So how would we differentiate ${\sin}^{n}x\,$ ?

T) $\frac{d}{dx}{\sin}^{n}x$ $=\frac{d}{dx}{\left(\sin x\right)}^{n}$ $=n{\left(\sin x\right)}^{n-1}\cdot \cos x$ $=n{\sin}^{n-1}x\cdot \cos x$
U) $\frac{d}{dx}{\cos}^{n}x$ $=\frac{d}{dx}{\left(\cos x\right)}^{n}$ $=n{\left(\cos x\right)}^{n-1}\cdot \left(-\sin x\right)$ $=-n{\cos}^{n-1}x\cdot \sin x$
V) $\frac{d}{dx}{\tan}^{n}x$ $=\frac{d}{dx}{\left(\tan x\right)}^{n}$ $=n{\left(\tan x\right)}^{n-1}\cdot {\sec}^{2}x$ $=n{\tan}^{n-1}x\cdot {\sec}^{2}x$

Note:

Examples

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

$f'\left(x\right)=5{\sin}^{4}x\cos x\,$
- Analyze Something power n
  - $f\left(x\right)={\sin}^{5}x$ $={\left(\sin x\right)}^{5}$
  - $f'\left(x\right)={\color{Red}5}$
  - $f'\left(x\right)=5{\color{Red}\left(\sin x\right)}$
  - $f'\left(x\right)=5{\left(\sin x\right)}^{{\color{Red}4}}$
  - $f'\left(x\right)=5{\left(\sin x\right)}^{4}{\color{Red}\left(\cos x\right)}$
  - $=5{\sin}^{4}x\cos x\,$
  - Alternatively,
    $f'\left(x\right)={\color{Red}5}$
    $f'\left(x\right)=5{\color{Red}\sin x}$
    $f'\left(x\right)=5{\sin}^{{\color{Red}4}}x$
    $f'\left(x\right)=5{\sin}^{4}x{\color{Red}\cos x}$

$f'\left(x\right)=-2\cos x\sin x\,$
- $={\left(\cos x\right)}^{2}$
  - $f'\left(x\right)={\color{Red}2}$
  - $f'\left(x\right)=2{\color{Red}\cos x}$
  - $f'\left(x\right)={\color{Red}-}2\cos x{\color{Red}\sin x}$

$f'\left(x\right)=6\tan x {\sec}^{2} x\,$
- $={\left(\tan x\right)}^{2}$
  - $f'\left(x\right)={\color{Red}2}$
  - $f'\left(x\right)=2{\color{Red}\tan x}$
  - $f'\left(x\right)=2\tan x{\color{Red}{\sec}^{2} x}$

$f'\left(x\right)=\frac{3 \sin x}{{\cos}^{4} x}$
- $={\left(\cos x\right)}^{-3}$
  - $f'\left(x\right)={\color{Red}-3}$
  - $f'\left(x\right)=-3{\color{Red}\left(\cos x\right)}$
  - $f'\left(x\right)=-3{\left(\cos x\right)}^{{\color{Red}-4}}$
  - $f'\left(x\right)=-3{\left(\cos x\right)}^{-4}{\color{Red}\left(-\sin x\right)}$
  - $=\frac{3 \sin x}{{\cos}^{4} x}$

$f'\left(x\right)=\frac{3}{2}\cos x\sqrt{\sin x}$
- $={\left(\sin x\right)}^{\frac{3}{2}}$
  - $f'\left(x\right)={\color{Red}\frac{3}{2}}$
  - $f'\left(x\right)=\frac{3}{2}{\color{Red}\left(\sin x\right)}$
  - $f'\left(x\right)=\frac{3}{2}{\left(\sin x\right)}^{{\color{Red}\frac{1}{2}}}$
  - $f'\left(x\right)=\frac{3}{2}{\left(\sin x\right)}^{\frac{1}{2}}{\color{Red}\cos x}$
  - $=\frac{3}{2}\cos x\sqrt{\sin x}$

More Products and Quotients

K) if $y=uv\,$ , where $u,v\,$ are functions of $x\,$

$\frac{dy}{dx}$ $=u\frac{dv}{dx}+v\frac{du}{dx}$

L) if $y=\frac{u}{v},$ where $u,v\,$ are functions of $x\,$

$\frac{dy}{dx}$ $=\cfrac{v\cfrac{du}{dx}-u\cfrac{dv}{dx}}{{v}^{2}}$

Note : $u,v\,$ can be any functions, including composite functions. We MUST apply the product/quotient formula before we start worrying about the composite functions. And checking is a MUST.

Examples

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

$\frac{dy}{dx}=\frac{e^{3x}\left(3-e^{4x}\right)}{\left(e^{4x}+1\right)^{2}}$
- Analyze :
  - $\frac{dy}{dx}=\frac{\qquad\qquad}{}$
  - Copy bottom (put brackets) $\frac{dy}{dx}=\frac{{\color{Red}\left(e^{4x}+1\right)}\qquad\qquad}{}$
  - Minus copy top $\frac{dy}{dx}=\frac{\left(e^{4x}+1\right)\cdot3e^{3x}-{\color{Red}e^{3x}}\qquad\qquad}{}$
  - Multiply differentiate bottom $\frac{dy}{dx}=\frac{\left(e^{4x}+1\right)\cdot3e^{3x}-e^{3x}{\color{Red}\left(4e^{4x}\right)}}{}$
  - Bottom square $\frac{dy}{dx}=\frac{\left(e^{4x}+1\right)\cdot3e^{3x}-e^{3x}\left(4e^{4x}\right)}{{\color{Red}\left(e^{4x}+1\right)^{2}}}$
  - $=\frac{e^{3x}\left(3e^{4x}+3-4e^{4x}\right)}{\left(e^{4x}+1\right)^{2}}$
  - $=\frac{e^{3x}\left(3-e^{4x}\right)}{\left(e^{4x}+1\right)^{2}}$

$\frac{dy}{dx}=\frac{2x}{x^{2}+1}\left(\cos 3x-\sin 2x \right)-\left(3\sin 3x +2\cos 2x\right)\ln\left(x^{2}+1\right)$
- Analyze
  - $\frac{dy}{dx}={\color{Red}\left(\cos 3x-\sin 2x \right)}$
  - $\frac{dy}{dx}=\left(\cos 3x-\sin 2x \right){\color{Red}\frac{}{x^{2}+1}}$
  - $\frac{dy}{dx}=\left(\cos 3x-\sin 2x \right)\frac{{\color{Red}2x}}{x^{2}+1}$
  - $\frac{dy}{dx}=\left(\cos 3x-\sin 2x \right)\frac{2x}{x^{2}+1}+{\color{Red}\ln\left(x^{2}+1\right)}$
  - $\frac{dy}{dx}=\left(\cos 3x-\sin 2x \right)\frac{2x}{x^{2}+1}+\ln\left(x^{2}+1\right){\color{Red}\left(-\quad\sin 3x \right.}$
  - $\frac{dy}{dx}=\left(\cos 3x-\sin 2x \right)\frac{2x}{x^{2}+1}+\ln\left(x^{2}+1\right)\left(-{\color{Red}3}\sin 3x \right.$
  - $\frac{dy}{dx}=\left(\cos 3x-\sin 2x \right)\frac{2x}{x^{2}+1}+\ln\left(x^{2}+1\right)\left(-3\sin 3x {\color{Red}-\quad\cos 2x}\right.$
  - $\frac{dy}{dx}=\left(\cos 3x-\sin 2x \right)\frac{2x}{x^{2}+1}+\ln\left(x^{2}+1\right)\left(-3\sin 3x -{\color{Red}2}\cos 2x\right)$

$\frac{dy}{dx}=\frac{2^{\sin x}\left(\cos 4x\cos x\ln 2+4\sin 4x \right)}{\left(\cos 4x \right)^{2}}$
- - $\frac{dy}{dx}=\frac{{\color{Red}\cos 4x}\qquad\qquad}{}$
  - $\frac{dy}{dx}=\frac{\cos 4x\cdot \qquad {\color{Red}2^{\sin x}\ln 2}\qquad\qquad}{}$
  - $\frac{dy}{dx}=\frac{\cos 4x\cdot {\color{Red}\left(\cos x \right)} 2^{\sin x}\ln 2\qquad\qquad}{}$
  - $\frac{dy}{dx}=\frac{\cos 4x\cdot \left(\cos x \right)2^{\sin x}\ln 2{\color{Red}-2^{\sin x}}}{}$
  - $\frac{dy}{dx}=\frac{\cos 4x\cdot \left(\cos x \right)2^{\sin x}\ln 2-2^{\sin x}{\color{Red}\left(-\quad\sin 4x \right)}}{}$
  - $\frac{dy}{dx}=\frac{\cos 4x\cdot \left(\cos x \right)2^{\sin x}\ln 2-2^{\sin x}\left(-{\color{Red}4}\sin 4x \right)}{}$
  - $\frac{dy}{dx}=\frac{\cos 4x\cdot \left(\cos x \right)2^{\sin x}\ln 2-2^{\sin x}\left(-4\sin 4x \right)}{{\color{Red}\left(\cos 4x \right)^{2}}}$

Exercise 5

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

a) $y={\left(3x+\sin x \right)}^{3}$ $\frac{dy}{dx}=3{\left(3x+\sin x \right)}^{2}\left(3+\cos x \right)$
b) $y=\frac{1}{\left(2e^{x}-1\right)^{4}}$ $\frac{dy}{dx}=-\frac{8e^{x}}{\left(2e^{x}-1\right)^{3}}$
c) $y=\sqrt{\sin x+\cos x}$ $\frac{dy}{dx}=\frac{\cos x-\sin x}{2\sqrt{\sin x+\cos x}}$
d) $y=\left(\ln x-5\sqrt{x}\right)^{2}$ $\frac{dy}{dx}=2\left(\ln x-5\sqrt{x}\right)\left(\frac{1}{x}-\frac{5}{2\sqrt{x}} \right)$
e) $y=\frac{1}{x^{2}-3\ln x}$ $\frac{dy}{dx}=\frac{3-2x^{2}}{x\left(x^{2}-3\ln x \right)^{2}}$
f) $y=\frac{3}{\sqrt{e^{x}+e^{-x}}}$ $\frac{dy}{dx}=\frac{3\left(e^{x}-e^{-x}\right)}{\left( 2e^{x}+e^{-x}\right)^\frac{3}{2}}$
g) $y=\frac{1}{e^{x}+4}$ $\frac{dy}{dx}=-\frac{e^{x}}{\left(e^{x}+4\right)^{2}}$
h) $y=\frac{1}{\tan x-\sin x}$ $\frac{dy}{dx}=\frac{\cos x-\sec^{2}x}{\left(\tan x-\sin x\right)^{2}}$

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

a) $f\left(x \right)=\cos^{4}x$ $f'\left(x \right)=-4\sin x\cos^{3}x$
b) $f\left(x \right)=\sin^{2}x$ $f'\left(x \right)=10\sin x\cos x$
c) $f\left(x \right)=-\frac{1}{2}\tan^{3}x$ $f'\left(x \right)=-\frac{3}{2}\tan^{2}x\sec^{2}x$
d) $f\left(x \right)=\frac{1}{\sin^{3}x}$ $f'\left(x \right)=-\frac{3\cos x}{\sin^{4}x}$
e) $f\left(x\right)=\sqrt{\sin^{5}x}$ $f'\left(x \right)=\frac{5}{2}\left(\sin x\right)^{\frac{3}{2}}\cos x$
f) $f\left(x\right)=\sqrt[3]{\sin^{2}x}$ $f'\left(x \right)=-\frac{2}{3}\left(\cos x\right)^{-\frac{1}{3}}\sin x$

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

a) $\frac{dy}{dx}=e^{-2x}\left(x+1\right)^{2}\left(1-2x\right)$
- $\begin{align} \frac{dy}{dx}&=e^{-2x}\cdot 3\left(x+1\right)^{2}+\left(x+1\right)^{3}\left( -2e^{-2x}\right)\\ &=e^{-2x}\left[3-2\left(x+1\right) \right]\\ &=e^{-2x}\left(x+1\right)^{2}\left(1-2x\right) \end{align}$
b) $y=\ln\left(3x-1\right)\sin\left(2x+5\right)$ $\frac{dy}{dx}=2\ln\left(3x-1\right)\cos\left(2x+5\right)+\frac{3\sin\left(2x+5\right)}{3x-1}$
c) $\frac{dy}{dx}=e^{3x}\left(5\sin 2x-\cos 2x \right)$
- $\begin{align} \frac{dy}{dx}&=e^{3x}\left(2\cos 2x+2\sin 2x \right)+\left(\sin 2x-\cos 2x \right)\cdot 3e^{3x}\\ &=e^{3x}\left(2\cos 2x+2\sin 2x+ 3\sin 2x-3\cos 2x\right)\\ &=e^{3x}\left(5\sin 2x-\cos 2x \right) \end{align}$
d) $\frac{dy}{dx}=x^{2}\left(2x\cos 2x+3\sin 2x \right)$
- $\begin{align} \frac{dy}{dx}&=x^{3}\left(2\cos 2x\right)+\left(\sin 2x\right)3x^{2}\\ &=x^{2}\left(2x\cos 2x+3\sin 2x \right) \end{align}$
e) $\frac{dy}{dx}=\frac{12e^{3x}}{\left(e^{3x}+2\right)^{2}}$
- $\begin{align} \frac{dy}{dx}&=\frac{\left(e^{3x}-2\right)\cdot 3e^{3x}-\left(e^{3x}+2\right)\cdot 3e^{3x}}{\left(e^{3x}+2\right)^{2}}\\ &=\frac{12e^{3x}}{\left(e^{3x}+2\right)^{2}} \end{align}$
f) $\frac{dy}{dx}=\frac{e^{2x-1}\left(2\sin3x-3\cos3x \right)}{\sin^{2} 3x}$
- $\begin{align} \frac{dy}{dx}&=\frac{\left(\sin 3x\right)2e^{2x-1}-e^{2x-1}\cdot3\cos 3x}{\sin^{2} 3x}\\ &=\frac{e^{2x-1}\left(2\sin3x-3\cos3x \right)}{\sin^{2} 3x} \end{align}$
g) $\frac{dy}{dx}=\frac{2\sin\left(x^{2}\right)-2x\left(2x+1\right)\ln\left(2x+1\right)\cos\left(x^{2}\right)}{\left(2x+1\right)\sin^{2}\left(x^{2}\right)}$
- $\begin{align} \frac{dy}{dx}&=\frac{\sin\left(x^{2}\right)\left(\cfrac{2}{2x+1}\right)-\ln\left(2x+1\right)\left[2x\cos\left(x^{2}\right)\right]}{\sin^{2} \left(x^{2}\right)}\\ &=\frac{2\sin\left(x^{2}\right)-2x\left(2x+1\right)\ln\left(2x+1\right)\cos\left(x^{2}\right)}{\left(2x+1\right)\sin^{2}\left(x^{2}\right)} \end{align}$

Multistage Composite Functions

When dealing composite functions, it is VERY IMPORTANT we identify what type of function it is. One way to see that is we say that we differentiate form out to in.

Looking at the last question from Exercise 4

$y=\sin\left({e}^{x}\right)+{e}^{\sin x}+\ln\left(\sin x\right)+\sin\left(\ln x\right)$
$y={\color{Red}\sin}\left({e}^{x}\right)+{{\color{Red}e}}^{\sin x}+{\color{Red}\ln}\left(\sin x\right)+{\color{Red}\sin}\left(\ln x\right)$
We can see the
- The first one is sin something
- The second one is e power something
- The third one is ln something
- The first one is sin something

Be very careful however if it is ${\sin}^{2}x\,$ .

When the composite function involve more that 2 stage, we can apply the same concept of out to in.

For example, to differentiate $\sin \left(e^{3x}\right)$ , there is in total 3 stage.

${\color{Red}\sin} \left(e^{3x} \right)$
Followed by $\sin \left({\color{Red}e}^{3x}\right)$
Followed by $\sin \left(e^{{\color{Red}3x}}\right)$

However, until the time you get comfortable to do it in one shot (and able to check it confidently), it's better to do it in two main steps only, moving the second and third as a side working. In other words

Identify the main function (something power n, e power something, ln something, sin something, etc)
If the something is another composite function, we differentiate it as side working.

Examples

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

$\frac{dy}{dx}=3e^{3x}\cos \left(e^{3x}\right)$
- Analyze sin something
  - Becomes cos
    $\frac{dy}{dx}={\color{Red}\cos}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=\cos\left(e^{3x}\right)\left(\qquad\qquad\right)\qquad\qquad\qquad & \cfrac{d}{dx}e^{3x}={\color{Red}3}e^{3x}\end{array}$
- Alternatively
  - $\frac{dy}{dx}={\color{Red}\cos}$
  - $\frac{dy}{dx}=\cos{\color{Red}\left(e^{3x}\right)}$
  - $\frac{dy}{dx}=\cos\left(e^{3x}\right){\color{Red}\left(\quad e^{3x}\right)}$
  - $\frac{dy}{dx}=\cos\left(e^{3x}\right)\left({\color{Red}3} e^{3x}\right)$

$\frac{dy}{dx}=3e^{\sin 3x}\cos 3x$
- Analyze e power something
  - $\frac{dy}{dx}={\color{Red}e^{\sin 3x}}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=e^{\sin 3x}{\color{Red}\left(\qquad\qquad\right)}\qquad\qquad\qquad & {\color{Red}\cfrac{d}{dx}\sin 3x}\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=e^{\sin 3x}\left(\qquad\qquad\right)\qquad\qquad\qquad & \cfrac{d}{dx}\sin 3x={\color{Red}\quad\cos 3x}\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=e^{\sin 3x}\left(\qquad\qquad\right)\qquad\qquad\qquad & \cfrac{d}{dx}\sin 3x={\color{Red}3}\cos 3x\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=e^{\sin 3x}\left({\color{Red}3\cos 3x}\right)\qquad\qquad\qquad & \cfrac{d}{dx}\sin 3x=3\cos 3x\end{array}$

$\frac{dy}{dx}=\frac{2\cos 2x}{\sin 2x}$
- Analyze
  - $\frac{dy}{dx}=\frac{}{{\color{Red}\sin 2x}}$
  - $\frac{dy}{dx}=\frac{\quad{\color{Red}\cos 2x}}{\sin 2x}$
  - $\frac{dy}{dx}=\frac{{\color{Red}2}\cos 2x}{\sin 2x}$

$\frac{dy}{dx}=6e^{2x}\left(e^{2x}+1 \right)^{2}$
- Analyze something power n
  - $\frac{dy}{dx}={\color{Red}3}$
  - $\frac{dy}{dx}=3{\color{Red}\left(e^{2x}+1\right)}$
  - $\frac{dy}{dx}=3\left(e^{2x}+1 \right)^{{\color{Red}2}}$
  - $\frac{dy}{dx}=3\left(e^{2x}+1 \right)^{2}{\color{Red}\left(\quad e^{2x}\right)}$
  - $\frac{dy}{dx}=3\left(e^{2x}+1 \right)^{2}\left({\color{Red}2}e^{2x}\right)$

$\frac{dy}{dx}=\frac{2e^{2x}+3\cos 3x}{2\sqrt{e^{2x}+\sin 3x}}$
- Analyze something power n
  - $\frac{dy}{dx}={\color{Red}\frac{1}{2}}$
  - $\frac{dy}{dx}=\frac{1}{2}{\color{Red}\left(e^{2x}+\sin 3x\right)}$
  - $\frac{dy}{dx}=\frac{1}{2}\left(e^{2x}+\sin 3x\right)^{{\color{Red}-\frac{1}{2}}}$
  - $\frac{dy}{dx}=\frac{1}{2}\left(e^{2x}+\sin 3x\right)^{-\frac{1}{2}}{\color{Red}\left(\quad e^{2x}\right.}$
  - $\frac{dy}{dx}=\frac{1}{2}\left(e^{2x}+\sin 3x\right)^{-\frac{1}{2}}\left({\color{Red}2}e^{2x}\right.$
  - $\frac{dy}{dx}=\frac{1}{2}\left(e^{2x}+\sin 3x\right)^{-\frac{1}{2}}\left(2e^{2x}{\color{Red}+\quad\cos 3x}\right.$
  - $\frac{dy}{dx}=\frac{1}{2}\left(e^{2x}+\sin 3x\right)^{-\frac{1}{2}}\left(2e^{2x}+{\color{Red}3}\cos 3x\right.{\color{Red}\left.\right)}$

$\frac{dy}{dx}=-2e^{x}\cos\left(e^{x}\right)\sin\left(e^{x}\right)$
- Analyze
  - $y={\cos}^{2}\left(e^{x} \right)$ $=\left[ \cos \left(e^{x}\right)\right]^{2}$
  - $\frac{dy}{dx}={\color{Red}2}$
  - $\frac{dy}{dx}=2{\color{Red}\left[\cos\left(e^{x}\right)\right]}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=2\left[\cos\left(e^{x}\right)\right]{\color{Red}\left(\qquad\qquad\right)}\qquad\qquad\qquad & {\color{Red}\cfrac{d}{dx}\cos\left(e^{x}\right)}\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=2\left[\cos\left(e^{x}\right)\right]\left(\qquad\qquad\right)\qquad\qquad\qquad & \cfrac{d}{dx}\cos\left(e^{x}\right)={\color{Red}-\quad\sin\left(e^{x}\right)}\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=2\left[\cos\left(e^{x}\right)\right]\left(\qquad\qquad\right)\qquad\qquad\qquad & \cfrac{d}{dx}\cos\left(e^{x}\right)=-{\color{Red}e^{x}}\sin\left(e^{x}\right)\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=2\left[\cos\left(e^{x}\right)\right]\left[{\color{Red}-e^{x}\sin\left(e^{x}\right)}\right]\qquad\qquad\qquad & \cfrac{d}{dx}\cos\left(e^{x}\right)=-e^{x}\sin\left(e^{x}\right)\end{array}$

$\frac{dy}{dx}=\frac{3x^{2}}{x^{3}+1}\cos\left[\ln\left(x^{3}+1\right)\right]$
- Analyze sin something
  - $\frac{dy}{dx}={\color{Red}\cos\left[\ln\left(x^{3}+1\right)\right]}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=\cos\left[\ln\left(x^{3}+1\right)\right]{\color{Red}\left(\qquad\qquad\right)}\qquad\qquad\qquad & {\color{Red}\cfrac{d}{dx}\ln\left(x^{3}+1\right)}\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=\cos\left[\ln\left(x^{3}+1\right)\right]\left(\qquad\qquad\right)\qquad\qquad\qquad & \cfrac{d}{dx}\ln\left(x^{3}+1\right)={\color{Red}\cfrac{}{x^{3}+1}}\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=\cos\left[\ln\left(x^{3}+1\right)\right]\left(\qquad\qquad\right)\qquad\qquad\qquad & \cfrac{d}{dx}\ln\left(x^{3}+1\right)=\cfrac{{\color{Red}3x^{2}}}{x^{3}+1}\end{array}$
  - $\begin{array}{ll} \cfrac{dy}{dx}=\cos\left[\ln\left(x^{3}+1\right)\right]\left({\color{Red}\cfrac{3x^{2}}{x^{3}+1}}\right)\qquad\qquad\qquad & \cfrac{d}{dx}\ln\left(x^{3}+1\right)=\cfrac{3x^{2}}{x^{3}+1}\end{array}$

$\frac{dy}{dx}=\frac{3x^{2}\cos\left(x^{3}+1 \right)}{\sin\left(x^{3}+1 \right)}$
- Analyze ln something
  - $\frac{dy}{dx}={\color{Red}\frac{}{\sin\left(x^{3}+1 \right)}}$
  - $\frac{dy}{dx}=\frac{{\color{Red}\quad\cos\left(x^{3}+1 \right)}}{\sin\left(x^{3}+1 \right)}$
  - $\frac{dy}{dx}=\frac{{\color{Red}3x^{2}}\cos\left(x^{3}+1 \right)}{\sin\left(x^{3}+1 \right)}$

$\frac{dy}{dx}=\frac{3}{x}{\sin}^{2}\left(\ln x \right)\cos\left(\ln x \right)$
- Analyze Something power n
  - $y={\sin}^{3}\left(\ln x \right)$ $=\left[\sin\left(\ln x \right)\right]^{3}$
  - $\frac{dy}{dx}={\color{Red}3}$
  - $\frac{dy}{dx}=3{\color{Red}\left[\sin\left(\ln x \right)\right]}$
  - $\frac{dy}{dx}=3\left[\sin\left(\ln x \right)\right]^{{\color{Red}2}}$
  - $\frac{dy}{dx}=3\left[\sin\left(\ln x \right)\right]^{2}{\color{Red}\cos\left(\ln x\right)}$
  - $\frac{dy}{dx}=3\left[\sin\left(\ln x \right)\right]^{2}\cos\left(\ln x \right){\color{Red}\left(\frac{1}{x}\right)}$

$\frac{dy}{dx}=3\left[\ln\left(\sin x\right)\right]^{2}\frac{\cos x}{\sin x}$
- Analyze Something power n
  - $\frac{dy}{dx}={\color{Red}3}$
  - $\frac{dy}{dx}=3{\color{Red}\left[\ln\left(\sin x\right)\right]}$
  - $\frac{dy}{dx}=3\left[\ln\left(\sin x\right)\right]^{{\color{Red}2}}$
  - $\frac{dy}{dx}=3\left[\ln\left(\sin x\right)\right]^{2}{\color{Red}\frac{}{\sin x}}$
  - $\frac{dy}{dx}=3\left[\ln\left(\sin x\right)\right]^{2}\frac{{\color{Red}\cos x}}{\sin x}$

$\frac{dy}{dx}=\frac{\sin\sqrt{x}\cos\sqrt{x}}{\sqrt{x}}$
- Analyze Something power n
  - $y=\sin^{2}\sqrt{x}$ $=\left(\sin \sqrt{x}\right)^{2}$
  - $\frac{dy}{dx}={\color{Red}2}$
  - $\frac{dy}{dx}=2{\color{Red}\sin\sqrt{x}}$
  - $\frac{dy}{dx}=2\sin \sqrt{x}{\color{Red}\cos \sqrt{x}}$
  - $\frac{dy}{dx}=2\sin \sqrt{x}\cos \sqrt{x}{\color{Red}\left(\frac{1}{2\sqrt{x}}\right)}$

$\frac{dy}{dx}=\frac{-e^{-x}\cos\left(e^{-x}+3 \right)}{2\sqrt{\sin\left(e^{-x}+3 \right)}}$
- Analyze Something power n
  - $\frac{dy}{dx}={\color{Red}\frac{1}{2}}$
  - $\frac{dy}{dx}=\frac{1}{2}{\color{Red}\left[ \sin\left(e^{-x}+3 \right)\right]}$
  - $\frac{dy}{dx}=\frac{1}{2}\left[ \sin\left(e^{-x}+3 \right)\right]^{{\color{Red}-\frac{1}{2}}}$
  - $\frac{dy}{dx}=\frac{1}{2}\left[ \sin\left(e^{-x}+3 \right)\right]^{-\frac{1}{2}}{\color{Red}\cos\left(e^{-x}+3 \right)}$
  - $\frac{dy}{dx}=\frac{1}{2}\left[ \sin\left(e^{-x}+3 \right)\right]^{-\frac{1}{2}}\cos\left(e^{-x}+3 \right){\color{Red}\left(-e^{-x}\right)}$

Exercise 6

Note : Make sure you have mastered all the examples here before attempting the exercises.

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

a) $y=e^{\sin 2x}\,$ $\frac{dy}{dx}=2e^{\sin 2x}\cos x$
b) $y=\sin\left(e^{-x}\right)$ $\frac{dy}{dx}=-e^{-x}\cos\left(e^{-x}\right)$
c) $y=\ln\left(\cos3x\right)$ $\frac{dy}{dx}=-\frac{3\sin 3x}{\cos 3x}$
d) $y=\left[\ln\left(\tan x\right)\right]^{4}$ $\frac{dy}{dx}=\frac{4\sec^{2}x\left[\ln\left(\tan x\right)\right]^{3}}{\tan x}$
e) $y=e^{\cos 4x}\,$ $\frac{dy}{dx}=-4e^{\cos 4x}\sin x$
f) $y=\ln\left(e^{3x^{2}}+2\right)$ $\frac{dy}{dx}=\frac{6xe^{3x^{2}}}{e^{3x^{2}}+2}$
g) $y=\sin\left(e^{4x-1}\right)$ $\frac{dy}{dx}=4e^{4x-1}\cos\left(e^{4x-1}\right)$
h) $y=\frac{1}{\sin\left(x^{2}\right)}$ $\frac{dy}{dx}=-\frac{2x\cos\left(x^{2}\right)}{\sin^{2}\left(x^{2}\right)}$
i) $y=\sin^{3}\left(e^{x}\right)$ $\frac{dy}{dx}=3e^{x}\sin^{2}\left(e^{x}\right)\cos\left(e^{x}\right)$
j) $y=\sqrt{\cos\left(e^{2x}\right)}$ $\frac{dy}{dx}=-\frac{e^{2x}\sin\left(e^{2x}\right)}{\sqrt{\cos\left(e^{2x}\right)}}$
k) $y=\cos^{2}\sqrt{x}$ $\frac{dy}{dx}=-\frac{\sin\sqrt{x}\cos\sqrt{x}}{\sqrt{x}}$
l) $y=\ln\left[\cos\left(\frac{1}{x}\right)\right]$ $\frac{dy}{dx}=\frac{\sin\left(\frac{1}{x}\right)}{x^{2}\cos\left(\frac{1}{x}\right)}$
m) $y=\left[\ln\left(\sin x\right)\right]^{5}$ $\frac{dy}{dx}=\frac{5\cos x\left[\ln\left(\sin x\right)\right]^{4}}{\sin x}$
n) $y=\tan^{4}\left(\ln x\right)$ $\frac{dy}{dx}=\frac{4\tan^{3}\left(\ln x\right)\sec^{2}\left(\ln x\right)}{x}$
o) $y=\sin\left[\ln\left(1-x^{2}\right)\right]$ $\frac{dy}{dx}=\frac{2x\cos\left[\ln\left(1-x^{2}\right)\right]}{x^{2}-1}$
p) $y=\sqrt{\ln\left(e^{3x}+4\right)}$ $\frac{dy}{dx}=\frac{3e^{3x}}{2\left(e^{3x}+4\right)\sqrt{\ln\left(e^{3x}+4\right)}}$