To find the derivative of the function, you must apply the product rule. The product rule is as follows

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

In this case, we have $\displaystyle f(x)=5x^2$ and $\displaystyle g(x)=\tan{x}$

Using the product rule from above, we have 

$\displaystyle f'=10x\tan{x}+5x^2\sec^2{x}$

Use the quotient rule to find the derivative.

$\displaystyle \frac{d}{dx}\frac{2x+1}{x^2}=\frac{x^2(2)-2x(2x+1)}{2x}$

$\displaystyle =\frac{2x^2-4x^2-2x}{2x}$

$\displaystyle =\frac{-2x^2-2x}{2x}$

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

Use the product rule to find the derivative.

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

To find the derivative of a sum, we take the derivative of each part of the sum independently. Additionally, we apply the rules of differentiation that tell us 

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

Differentiating the function from the problem statement, we get

$\displaystyle f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}(4x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(10x)$

$\displaystyle f'(x)=12x^2+10$

Using the product and quotient rules to solve the problem, we get

$\displaystyle f'(x)=\frac{\cos(x)*\frac{\mathrm{d} }{\mathrm{d} x}(5x\tan(x))-(5x\tan(x))*\frac{\mathrm{d} }{\mathrm{d} x}(\cos(x))}{\cos^2(x)}$

The product rule is used when taking $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x\tan(x))=\frac{\mathrm{d} }{\mathrm{d} x}(5x)*\tan(x)+5x*\frac{\mathrm{d} }{\mathrm{d} x}\tan(x)$

The final answer is

$\displaystyle f'(x)=(\cos(x)\left (5\tan(x)+5x\sec^2(x))+5x\tan(x)\sin(x))/\cos^2(x)$

Using the product rule, we ge$\displaystyle f'(x)=8x^3-3x^2+6$t

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

Simplifying, 

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

$\displaystyle f'(x)=6x^3-3x^2+2x^3+6$

$\displaystyle f'(x)=8x^3-3x^2+6$

So whenever you have two distinct functions that are multiplied by each other, you will be using the product rule. So when looking at a function, see if you can separate it into two. In this case, we can see there is the function:

$\displaystyle 5x^3$ and the function $\displaystyle e^x$.

So lets call those $\displaystyle f(x) and g(x)$

Then the product rule is that the derivative of $\displaystyle f(x)g(x )$ is:

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

Then calculate to find:

$\displaystyle f'(x)=15x^2$ and $\displaystyle g'(x)=e^x$ to give an answer of:

$\displaystyle f'(x)g(x)+f(x)g'(x)=15x^2e^x+5x^3e^x$

which simplifies to:

$\displaystyle 5x^2e^x(3+x)$

To find the derivative of the sum, we take the derivative of each term independently, then add them all up. Further, we use the rule

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^n)=nx^{n-1}$

$\displaystyle f'=\frac{\mathrm{d} }{\mathrm{d} x}(2x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(5x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(x)+\frac{\mathrm{d} }{\mathrm{d} x}(-4)$

$\displaystyle f'=6x^2+10x+1$

To find the derivative of the function, we must use the product rule, which is

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

Using the function from the problem statement, we get

$\displaystyle f'=\frac{\mathrm{d} }{\mathrm{d} x}(5x)*(\cos(x))+\frac{\mathrm{d} }{\mathrm{d} x}(\cos(x))*5x$

$\displaystyle f'=5\cos(x)-5x\sin(x)$

To find the derivative of the function, we use the quotient rule, which is

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{f}{g})=\frac{gf'-fg'}{g^2}$, where $\displaystyle f$ and $\displaystyle g$ are any expression

Using the function from the problem statement, we get

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

Taking the derivatives, we get

$\displaystyle f'=\frac{\tan(x)(12x^2+\cos(x))-(4x^3+\sin(x))(\sec^2(x))}{\tan^2(x)}$