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