To find the derivative of this function, $\displaystyle f(x)=cos(x^3+2x^2+x+3)$, utilize the chain rule:

$\displaystyle cos(g(x))'=-g'(x)sin(g(x))$

Where

$\displaystyle g(x)=x^3+2x^2+x+3$

Performing the derivative gives:

$\displaystyle g'(x)=3x^2+4x+1$

So the derivative is:

$\displaystyle f'(x)=-(3x^2+4x+1)sin(x^3+2x^2+x+3)$

For the function $\displaystyle f(x)=\frac{e^{x^2}}{ln(x)}$, it will be useful to utilize the chain rule and quotient rule of derivatives.

The quotient rule states:

$\displaystyle \frac{d}{dx}\frac{u}{v}=\frac{vdu-udv}{v^2}$

Thus

$\displaystyle f'(x)=\frac{ln(x)(e^{x^2})'-e^{x^2}(ln(x))'}{ln^2(x)}$

The $\displaystyle ln(x)'$ value is simply $\displaystyle \frac{1}{x}$.

To find the value of $\displaystyle e^{x^2}'$, note that it should follow the form

$\displaystyle e^{g(x)}=g'(x)e^{g(x)}$

$\displaystyle e^{x^2}'=2xe^{x^2}$

Putting everything together:

$\displaystyle f'(x)=\frac{2xe^{x^2}ln(x)-\frac{1}{x}e^{x^2}}{ln^2(x)}$

$\displaystyle f'(x)=\frac{2x^2e^{x^2}ln(x)-e^{x^2}}{xln^2(x)}$

To solve this problem, we must use the following formulae:

$\displaystyle \frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x), \frac{d}{dx}(cx^{n})=ncx^{n-1}$

$\displaystyle \frac{d}{dx}\cos(x)=-\sin(x)$

Where $\displaystyle f(x)=3x^{2}$ and $\displaystyle g(x)=\cos(x)$

$\displaystyle \frac{d}{dx}(3x^{2}\cos(x))=3*2x^{2-1}\cos(x)-3x^{2}\sin(x)=6x\cos(x)-3x^{2}\sin(x)$

To find this derivative, we need to use the following formulae: 

$\displaystyle \frac{d}{dx}(\frac{g(x)}{h(x)})=\frac{g'(x)h(x)-g(x)h'(x)}{g(x)^{2}}$

$\displaystyle \frac{d}{dx} \sin(x)= \cos(x), \frac{d}{dx} \ln(x)=\frac{1}{x}, \frac{d}{dx}(c)=0$

Where $\displaystyle g(x)=\ln(x)$ and $\displaystyle h(x)=1+\sin(x)$.

$\displaystyle \frac{\frac{1}{x}*(1+sin(x))-cos(x)ln(x))}{(sin(x)+1)^{2}}$$\displaystyle =\frac{\frac{1+sin(x)}{x}-cos(x)ln(x)}{sin^{2}(x)+2sin(x)+1}$

To find the derivative of $\displaystyle f(x)$, we need the following formulae:

$\displaystyle \frac{d}{dx} \csc(x)=-\csc(x)\cot(x), \frac{d}{dx} f(g(x))=f'(g(x))g'(x)$

$\displaystyle \frac{d}{dx} a^{x}=a^{x}\ln(a)$

$\displaystyle f'(x)=\ln(3)*3^{\csc(x)}*-\csc(x)\cot(x)$

To find the derivative of f(x), we must use the following formulae:

$\displaystyle \\ \frac{d}{dx}(\log{_{a}}(x))=\frac{1}{x\ln(a)},\\ \frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)$

In this particular case,

$\displaystyle \\a=4, \\ g(x)=x^3$

thus our derivative becomes,

$\displaystyle f'(x)= \frac{1}{x^{3}ln(4)}*3x^{2} = \frac{3}{xln(4)}$.

To find the first derivative of $\displaystyle f(x)$, we must use the following formulae:

$\displaystyle \frac{d}{dx} \ln(x)=\frac{1}{x}, \frac{d}{dx} f(g(x))=f'(g(x))g'(x)$ 

Applying these rules we can find the following derivative.

Let,

$\displaystyle \\f(x)=\frac{1}{2x^2}\\ \\g(x)=2x^2$

therefore we get,

$\displaystyle f'(x)= \frac{1}{2x^{2}}*4x= \frac{2}x{}$.

To find this derivative, we need to use the following formulae:

$\displaystyle \\ \frac{d}{dx}(x^{n}))=nx^{n-1}, \\ \\ \frac{d}{dx} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

$\displaystyle \\ \frac{d}{dx} \cos(x)= -\sin(x),\\ \\ \frac{d}{dx} f(g(x))=f'(g(x))g'(x)$

Applying these rules where 

$\displaystyle \\ f(x)=x^4 \\ \\ g(x)=\cos(2x)$

therefore we get,

$\displaystyle f'(x)=4x^{3}\cos(2x)-x^{4}\sin(2x)*2$.

To find this derivative, we must use the product rule, power rule, chain rule, and trigonometric derivative rule for tangent.

Lets recall the product rule,

$\displaystyle \frac{d}{dx} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

In this particular problem,

$\displaystyle f(x)=3x^2$ and $\displaystyle g(x)=tan(7x)$.

In order to find $\displaystyle f'(x)$ we will need to use the power rule which states, 

$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$.

Therefore,

$\displaystyle f'(x)=2\cdot 3x^{2-1}=6x$.

To find $\displaystyle g'(x)$ we need to use the chain rule which states,

$\displaystyle \frac{d}{dx} h(i(x))=h'(i(x))i'(x)$

where $\displaystyle h(x)=tan(7x)$ and $\displaystyle i(x)=7x$.

To find $\displaystyle h'(x)$ we will need to use the trigonometric derivative rule for tangent which states,

$\displaystyle \frac{d}{dx}tan(x)=sec^2(x)$ and to find $\displaystyle i'(x)$ we will again use the power rule.

Thus,

$\displaystyle h'(x)=sec^2(7x)$ and $\displaystyle i'(x)=7\cdot 1x^{1-1}=7$.

This then makes,

$\displaystyle g'(x)=7sec^2(7x)$.

Now lets combine our terms using the product rule to find the final derivative.

$\displaystyle \\f'(x)=6x\tan(7x)+3x^{2}\cdot 7sec^{2}(7x) \\ f'(x)= 6x\tan(7x)+21x^{2}sec^{2}(7x)$

This problem relies on the chain rule

$\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))g'(x)$.

First, you have x to something, so you must use the power rule

$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$

to get 

$\displaystyle (e^{-2x})'x^{e^{-2x}-1}$.

Using the chain rule, this becomes $\displaystyle -2e^{-2x}x^{e^{-2x}-1}$.