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

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

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

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

\(\displaystyle f'=\frac{\mathrm{d} }{\mathrm{d} x}(12x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(\sin(x))-\frac{\mathrm{d} }{\mathrm{d} x}(100)\)

\(\displaystyle f'=24x+\cos(x)\)

To find the derivative of the function, we take the derivative of each element in the function independently, then add them up.

Using \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}\), we solve

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(5x)+\frac{\mathrm{d} }{\mathrm{d} x}(-1)=4x+5\)

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

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(f*g)=(f'*g)+(f*g')\), where f and g are both functions.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^5\sin{x})=\frac{\mathrm{d} }{\mathrm{d} x}(2x^5)*(\sin{x})+\frac{\mathrm{d} }{\mathrm{d} x}(\sin{x})*(2x^5)=10x^4\sin{x}+2x^5\cos{x}\)

To find the derivative of the function using the quotient rule, we apply the following definition:

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

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{e^x}{\tan{x}})=\frac{\tan{x}*\frac{\mathrm{d} }{\mathrm{d} x}(e^x)-e^x*\frac{\mathrm{d} }{\mathrm{d} x}(\tan{x})}{\tan^2{x}}=\frac{e^x\tan{x}-e^x\sec^2{x}}{\tan^2{x}}\)

To find the derivative of the function, we take the derivative of each element in the function independently, then add them up.

Using \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}\), we solve

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(5x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(-1)=30x^2+10x\)

Finding our second derivative requires two steps, we first must find the derivative then find the corresponding rate of change for that new equation.

\(\displaystyle f(x)=ln(x^2-1)\)

Here, the chain rule is used since our function is of the form \(\displaystyle [f(g(x))]dx=f'(x)g'(f(x))\)

\(\displaystyle f'(x)=\frac{2x}{x^2-1}\)

We now must use the quotient rule since our function is a rational function. We use the rule \(\displaystyle [\frac{f(x)}{g(x)}]dx=\frac{f'(x)g(x)-g'(x)f(x))}{(g(x))^2}\)

Therefore,

\(\displaystyle f''(x)=\frac{-2x^2-2}{x^4-2x^2+1}\)