$\displaystyle y =x^2\sin(3x)$

Differentiate both sides and proceed with the product rule: 

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

$\displaystyle \frac{dy}{dx}= x^2\frac{d}{dx}\sin(3x)+ \sin(3x)\frac{d}{dx}x^2$

Evaluate the derivatives in each term. For the first term,  

$\displaystyle =x^2\underbrace{\frac{d}{dx}\sin(3x)} + \sin(3x)\frac{d}{dx}x^2$                                                (2)

 apply the chain rule, 

$\displaystyle \frac{d}{dx}sin(3x)=3\cos(3x)$

So now the first term in equation (2) can be written, 

$\displaystyle x^2\frac{d}{dx}\sin(3x)=x^2[3\cos(3x)]=3x^2\cos(3x)$                            (3)

The second term in equation (2) is easy, this is just the product of  $\displaystyle \sin(3x)$multiplied by the derivative of $\displaystyle x^2$, 

$\displaystyle \sin(3x)\frac{d}{dx}x^2 = 2x\sin(3x)$                                                               (4)

Combine equations (3) and (4) to write the derivative, 

$\displaystyle \frac{dy}{dx} =3x^2\cos(3x)+2x\sin(3x)$

Use the product rule to find the derivative. 

$\displaystyle \cos (x)x^2+2x\sin (x)$

Use the power rule to find the derivative.

$\displaystyle \frac{d}{dx}x^3=3x^2$

$\displaystyle \frac{d}{dx}4x^2=8x$

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

Thus, the derivative is $\displaystyle 3x^2+8x-1$

Here we use the product rule: $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(F(x)\cdot G(x))=F'(x)\cdot G(x)+F(x)\cdot G'(x)$

Let $\displaystyle F(x)=3sin(x^2)$ and $\displaystyle G(x)=e^{2x-1}$

Then $\displaystyle F'(x)=3cos(x^2)\cdot2x$ (using the chain rule)

and $\displaystyle G'(x)=e^{2x-1}\cdot2$ (using the chain rule)

Subbing these values back into our equation gives us

$\displaystyle f'(x)=3cos(x^2)\cdot2x\cdot e^{2x-1}+3sin(x^2)\cdot e^{2x-1}\cdot2$

Simplify by combining like-terms

$\displaystyle f'(x)=6xcos(x^2)\cdot e^{2x-1}+6sin(x^2)\cdot e^{2x-1}$

and pulling out a $\displaystyle 6e^{2x-1}$ from each term gives our final answer

$\displaystyle f'(x)=6e^{2x-1}(xcos(x^2)+sin(x^2))$

When evaluating the derivative, pay attention to the fact that $\displaystyle e,\pi$ are constants, (not variables) and are treated as such.

$\displaystyle f'(x)=\pi\cos(ex)\times (ex)'=e\pi\cos(ex)$.

and hence

$\displaystyle f'(0)=e\pi\cos(e(0))=e\pi$.

To obtain an expression for $\displaystyle f'(x)$, we can take the derivative of $\displaystyle f(x)$ using the sum rule.

$\displaystyle f'(x)=0+1+2x+3x^2+4x^3$.

Substituting $\displaystyle 2$ into this equation gives us

$\displaystyle f'(2) =0+1+4+12+32$

$\displaystyle =49$.

To find $\displaystyle f'(x)$, we will need to use the quotient rule; $\displaystyle \frac{gf'-fg'}{g^2}$.

$\displaystyle f(x)= \frac{x\tan^{-1}x}{\ln(x)}$. Start

$\displaystyle f'(x)= \frac{(\ln(x))(x\tan^{-1}x)'-(x\tan^{-1}x)(\ln(x))'}{(\ln(x))^2}$. Use the quotient rule.

$\displaystyle = \frac{(\ln(x))(x(\frac{1}{1+x^2})+1(\tan^{-1}x))-(x\tan^{-1}x)(\frac{1}{x})}{(\ln(x))^2}$. Take the derivatives inside of the quotient rule. The derivative of $\displaystyle x\tan^{-1}x$ uses the product rule.

$\displaystyle =\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$. Simplify to match the correct answer.

Because this problem contains two functions multiplied together that can't be simplified any further, it calls for the product rule, which states that $\displaystyle \frac{d}{dx}(f(x)g(x)) = f{}'(x)g(x) + f(x)g{}'(x)$.

By using this rule, we get the answer:

$\displaystyle \cos(x)\cos(x) + \sin(x)(-\sin (x))$

By simplifying, we conclude that the derivative is equal to 

$\displaystyle \cos ^{2}(x)-\sin ^{2}(x)$.

$\displaystyle \begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }cos(19x)\\&\text{and the second being: }\frac{1}{x^{20}}\\&\text{With the product scaled by }-114\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[cos(ax)]=-asin(ax)dx\\&d[x^a]=ax^{a-1}dx\\&\text{From the above we can continue:}\\&u=cos(19x);du=-19sin(19x)\\&v=\frac{1}{x^{20}};dv=-\frac{20}{x^{21}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(2280cos(19x))}{x^{21}}+\frac{ (2166sin(19x))}{x^{20}}\end{align*}$

$\displaystyle \begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }tan(17x)\\&\text{and the second being: }\frac{1}{2^{(20x)}}\\&\text{With the product scaled by }-380\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&\text{From the above we can continue:}\\&u=tan(17x);du=17tan(17x)^{2} + 17\\&v=\frac{1}{2^{(20x)}};dv=-\frac{(20ln(2))}{2^{(20x)}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(7600tan(17x)ln(2))}{2^{(20x)}}-\frac{ (380\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}\end{align*}$