Computation of the derivative requires the use of the Product Rule and Chain Rule. 

The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:

$\displaystyle f(x)*g(x)$

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

This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."

In the problem statement, we are given:

$\displaystyle f(x)=3x^2(x^2-5x+2)^5$

$\displaystyle 3x^2$ is the "First" function, and $\displaystyle (x^2-5x+2)^5$ is the "Second" function. 

The "Second" function requires use of the Chain Rule. 

When:

$\displaystyle y=(f(x))^a \rightarrow \frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Applying these formulas results in:

$\displaystyle f'(x)=(3x^2)[5(x^2-5x+2)^4(2x-5)]+(x^2-5x+2)^5(6x)$

Simplifying the terms inside the brackets results in:

$\displaystyle f'(x)=(3x^2)[(10x-25)(x^2-5x+2)^4]+(x^2-5x+2)^5(6x)$

We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".

$\displaystyle f'(x)=(x^2-5x+2)^4[(3x^2)(10x-25)+(6x)(x^2-5x+2)]$

Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:

$\displaystyle f'(x)=(x^2-5x+2)^4[30x^3-75x^2+6x^3-30x^2+12x]$

Simplifying this results in one of the answer choices:

$\displaystyle f'(x)=(x^2-5x+2)^4(36x^3-105x^2+12x)$

Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of $\displaystyle \tan^{-1}(x)$.

Product Rule:

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

$\displaystyle \frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

Applying these two rules results in:

$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[x^2\tan^{-1}(3x^5)]$

$\displaystyle \frac{dy}{dx}=(x^2)(\frac{1}{1+(3x^5)^2})(15x^4)+\tan^{-1}(3x^5)(2x)$

$\displaystyle \frac{dy}{dx}=\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)$

This matches one of the answer choices.

Computation of this derivative will require the use of the Product Rule, and knowledge of the derivative of the inverse tangent function, and natural logarithmic function:

$\displaystyle \frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

$\displaystyle \frac{d}{dx}[\ln(u)]=\frac{1}{u}\frac{du}{dx}$

We can now easily compute the derivative.

$\displaystyle f'(x)=(2x^2)(\frac{1}{1+(3x)^2})(3)+(\tan^{-1}(3x))(4x)+5(\frac{1}{10x})(10)$

This simplifies to:

$\displaystyle f'(x)=\frac{6x^2}{1+9x^2}+4x\tan^{-1}(3x)+\frac{5}{x}$

This is one of the answer choices.

Solving for the derivative requires knowledge of the rule for the inverse tangent function:

$\displaystyle \frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

In our case:

 $\displaystyle u=2x(x+1)^{10}$

We can take the derivative of this using the product rule:

$\displaystyle \frac{du}{dx}=(2x)(10(x+1)^9(1))+(2)(x+1)^{10}=2(x+1)^9[10x+(x+1)]$

$\displaystyle \frac{du}{dx}=2(x+1)^9(11x+1)$

Now we can simply plug all of this into the above formula and we arrive at:

$\displaystyle \frac{dy}{dx}=[\frac{1}{1+(2x(x+1)^{10})^2}][2(x+1)^9(11x+1)]$

Simplifying this further gives:

$\displaystyle \frac{dy}{dx}=2(x+1)^{9}[\frac{11x+1}{1+4x^2(x+1)^{20}}]$

In order to find the slope of a function at a certain point, plug in that point into the first derivative of the function. Our first step here is to take the first derivative.

Since we see that f(x) is composed of two different functions, we must use the product rule. Remember that the product rule goes as follows:

$\displaystyle f(x)=g(x)h(x)$

$\displaystyle f'(x)=g'(x)h(x) + g(x)h'(x)$

Following that procedure, we set $\displaystyle ln(x)$ equal to $\displaystyle g(x)$ and $\displaystyle x^2$ equal to $\displaystyle h(x)$.

$\displaystyle f'(x)= \frac{1}{x}(x^2)+ ln(x)(2x)$,

which can be simplified to

$\displaystyle f'(x)=x+2xln(x) = x(1+2ln(x))$.

Now plug in 1 to find the slope at x=1.

$\displaystyle f'(1)= 1(1+2(0))=1$

Remember that $\displaystyle ln(1)=0$.

To solve this problem, first, we need to take the derivative of the function. To do this we need to use the quotient rule and simplified as follows:

 $\displaystyle \\f'(x)=\frac{(x+4)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2)-(x^2+2)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+4)}{(x+4)^2}\\ f'(x)=\frac{(x+4)\cdot 2x-(x^2+2)\cdot 1}{(x+4)^2}\\f'(x)=\frac{2x^2+8x-x^2-2}{(x+4)^2}\\f'(x)=\frac{x^2+8x-2}{(x+4)^2}$

From here we need to evaluate at the given point $\displaystyle (2,1)$. In this case, only the x value is important, so we evaluate our derivative at $\displaystyle x=2$ to get

$\displaystyle \\f'(2)=\frac{(2)^2+8(2)-2}{(2+4)^2}\\ f'(2)=\frac{4+16-2}{36}\\f'(2)=\frac{18}{36}=\frac{1}{2}$

To find the second derivative, first we need to find the first derivative. To find the first derivative we need to use the quotient rule as follows. So for the given function, we get the first derivative to be 
$\displaystyle \\f'(x)=\frac{x\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2x+1)-(x^2+2x+1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x)}{x^2}\\f'(x)=\frac{x\cdot (2x+2)-(x^2+2x+1)\cdot 1}{x^2}\\f'(x)=\frac{x^2-1}{x^2}$

Now we have to take the derivative of the derivative. To do this we need to use the quotient rule as shown below.

 Thus, we get 

$\displaystyle \\f''(x)=\frac{x^2\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2-1)-(x^2-1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2)}{x^4}\\ f''(x)=\frac{2x^3-2x^3+2x}{x^4}\\f''(x)=\frac{2x}{x^4}=\frac{2}{x^3}$

We find the answer using the quotient rule

$\displaystyle (\frac{f(x)}{g(x)})' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

and the product rule

$\displaystyle (f(x)g(x))' = f(x)g'(x)+g(x)f'(x)$

and then simplifying.

$\displaystyle y' = \frac{x\ln(x)\times 2x-x^2\times(x\frac{1}{x}+\ln(x)))}{(x\ln(x))^2}$

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

$\displaystyle =\frac{2\ln(x)-1-\ln(x)}{\ln(x)^2}$

$\displaystyle =\frac{\ln(x)-1}{\ln(x)^2}$ or $\displaystyle \frac{\ln(x)-1}{[\ln(x)]^2}$.

The extra brackets in the denominator are optional.

Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have

$\displaystyle x^y=yx$

$\displaystyle \ln(x^y)=\ln(xy)$

$\displaystyle y\ln(x)=\ln(x)+\ln(y)$

$\displaystyle y = \frac{\ln(y)}{\ln(x)}+1$

$\displaystyle y' = \frac{\ln(x)\frac{y'}{y}-\frac{\ln(y)}{x}}{\ln(x)^2}$. Quotient Rule + implicit differentiation.

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

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

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

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

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

$\displaystyle y'=\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

The correct answer is $\displaystyle \frac{x-(1+x^2)\tan^{-1}(x)}{x^2(1+x^2)}$.

Using the Quotient Rule and the fact $\displaystyle \frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}$, we have

$\displaystyle y = \frac{\tan^{-1}(x)}{x}$

$\displaystyle y' = \frac{x\frac{1}{1+x^2}-\tan^{-1}(x)}{x^2}$

$\displaystyle y' = \frac{\frac{x}{1+x^2}-\tan^{-1}(x)}{x^2}$

$\displaystyle y' = \frac{x}{x^2(1+x^2)}-\frac{\tan^{-1}(x)}{x^2}$

$\displaystyle y' = \frac{x}{x^2(1+x^2)}-\frac{\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}$

$\displaystyle y' = \frac{x-\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}$