Take the derivative of each term.

$\displaystyle \tfrac{\mathrm{d} }{\mathrm{d} x}(15x^5)=75x^4$

$\displaystyle \tfrac{\mathrm{d} }{\mathrm{d} x}(-2x^4)=-8x^3$

$\displaystyle \tfrac{\mathrm{d} }{\mathrm{d} x}(11x^2)=22x$

$\displaystyle \tfrac{\mathrm{d} }{\mathrm{d} x}(e^x)=e^x$

$\displaystyle \tfrac{\mathrm{d} }{\mathrm{d} x}(-\ln x)=-\tfrac{1}{x}$

Add them:

$\displaystyle 75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

There are two ways to solve this problem.

First, you can use a trig identity to replace $\displaystyle 2\sin x\cos x$ with $\displaystyle \sin 2x$. Using the chain rule, $\displaystyle (\sin 2x)'=2\cos2x$.

Alternatively, you could use the product rule.

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

Since $\displaystyle \cos^2x-\sin^2x=\cos2x$, our final answer is still $\displaystyle 2\cos2x$.

For this problem, we need to use the quotient rule.

$\displaystyle (\frac{f}{g})'=\frac{gf'-fg'}{g^2}$

$\displaystyle (\frac{e^x}{\sin x})'=\frac{\sin x(e^x)-e^x(\cos x)}{\sin^2x}$

Simplifying:

$\displaystyle e^x (\frac{\sin x-\cos x}{\sin^2x})=\frac{e^x-e^x\tan x}{\sin x}$

$\displaystyle \frac{d}{dx}[2x^4+5x]=8x^3+5$

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

Product rule states: $\displaystyle \frac{d}{du}[f(u) \cdot g(u)]=f'(u)g(u)+g'(u)f(u)$

Therefore:

 $\displaystyle y'=(8x^3+5)(cos(x))-sin(x)(2x^4+5x)$

$\displaystyle =-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)$

$\displaystyle h(x)=sin(x^{2}e^{x})$

By the chain rule: $\displaystyle h^{'}(x)=cos(x^{2}e^{x})\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$

Differentiate $\displaystyle (x^{2}e^{x})$ using the product rule: $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})e^{x}+x^{2}\frac{\mathrm{d} }{\mathrm{d} x}e^{x}=2xe^{x}+x^{2}e^{x}$

Substitute this derivative for $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$ in the first equation: $\displaystyle h^{'}(x)=cos(x^{2}e^{x})(2xe^{x}+x^{2}e^{x})$

Factor the equation: $\displaystyle h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

$\displaystyle \frac{d}{dx}[cos(2p-1)]=2 \cdot -sin(2p-1)=-2sin(2p-1)$

$\displaystyle \frac{d}{dx}[2p^3+4p]=6p^2+4$

Therefore:

$\displaystyle f'(p)=-2sin(2p-1)-(6p^2+4)$

$\displaystyle =-2sin(2p-1)-6p^2-4$

$\displaystyle =-2(sin(2p-1)+3p^2+2)$

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

The second derivative is just the derivative of the first derivative. So first we find the first derivative of $\displaystyle f(x)$. Remember the derivative of $\displaystyle \sin x$ is $\displaystyle \cos x$, and the derivative for $\displaystyle \cos x$ is $\displaystyle -\sin x$.

 $\displaystyle f'(x)= cos(x)+sin(x)$

Then to get the second derivative, we just derive this function again. So

$\displaystyle f''(x)=-sin(x)+cos(x)=cos(x)-sin(x)$

Take the derivative $\displaystyle f'$ of $\displaystyle f$, then take the derivative of $\displaystyle f'$.

$\displaystyle f (x) = 6x^{3} - 12x^{2} + 4x -8$

$\displaystyle f '(x) = 3 \cdot 6x^{3-1} - 2 \cdot 12x^{2-1} + 4 -0$

$\displaystyle f '(x) = 18x^{2} - 24x+ 4$

$\displaystyle f ' '(x) = 2 \cdot 18x^{2-1} - 24+ 0$

$\displaystyle f ' '(x) = 36x - 24$

Take the derivative $\displaystyle g'$ of $\displaystyle g$, then take the derivative of $\displaystyle g'$.

$\displaystyle g(x) = \cos \left (x^{2} \right )$

$\displaystyle g'(x) = 2x \cdot [-\sin (x^2)]$

$\displaystyle g'(x) = - 2x \sin (x^2)$

$\displaystyle g''(x) = - 2\left [ 1 \cdot \sin(x^2)+ x \cdot 2x \cos(x^2) \right ]$

$\displaystyle g''(x) = - 2\left [ \sin(x^2)+ 2x^2 \cos(x^2) \right ]$

$\displaystyle g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

Rewrite:

$\displaystyle f(x) = \frac{1}{e^{4x}}$

$\displaystyle f(x) = e^{-4x}$

Take the derivative $\displaystyle f'$ of $\displaystyle f$, then take the derivative of $\displaystyle f'$.

$\displaystyle f'(x) = -4 e^{-4x}$

$\displaystyle f''(x) = -4 \cdot \left (-4 \right )e^{-4x} = 16e^{-4x} = \frac{16 }{e^{4x}}$