The derivative of the function is equal to

\(\displaystyle f'(x)=3\sec(3x)\tan(3x)+4xe^{2x^2+1}\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)\)

In order to solve for the first and second derivative, we must use the chain rule.

The chain rule states that if

\(\displaystyle y=f(u)\) 

and 

\(\displaystyle u=g(x)\)

then the derivative is

\(\displaystyle \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}\)

In order to find the first derviative of the function

\(\displaystyle f(x)=e^{\pi x}\)

we set

\(\displaystyle y=e^u\)

and

\(\displaystyle u=\pi x\)

Because the derivative of the exponential function is the exponential function itself, we get

\(\displaystyle \frac{dy}{du}=\frac{d}{du}[e^u]\)

\(\displaystyle \frac{dy}{du}=e^u\)

And differentiating \(\displaystyle u\) we use the power rule which states

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

As such

\(\displaystyle \frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]\)

\(\displaystyle \frac{du}{dx}=\pi x^{1-1}=\pi\)

And so

\(\displaystyle \frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}\)

\(\displaystyle f'(x)=\pi e^{\pi x}\)

To solve for the second derivative we set 

\(\displaystyle y=\pi e^{u}\)

and 

\(\displaystyle u=\pi x\)

Because the derivative of the exponential function is the exponential function itself, we get

\(\displaystyle \frac{dy}{du}=\frac{d}{du}[\pi e^u]\)

\(\displaystyle \frac{dy}{du}=\pi e^u\)

And differentiating \(\displaystyle u\) we use the power rule which states

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

As such

\(\displaystyle \frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]\)

\(\displaystyle \frac{du}{dx}=\pi x^{1-1}=\pi\)

And so the second derivative becomes

\(\displaystyle \frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}\)

\(\displaystyle f''(x)=\pi^2e^{\pi x}\)

The first derivative of the function is equal to

\(\displaystyle f'(x)=3x^2\cos(x)-x^3\sin(x)+2e^{2x}\)

The second derivative - the derivative of the function above - of the original function is equal to

\(\displaystyle f''(x)=\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}\)

Both derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}\)

We have a function inside of another function, therefore we have a chain-rule.

\(\displaystyle f' (x) =4 \left( x^3 - \sin {x} \right ) ^3 \left( 3x ^ 2 - \cos x \right )\).

At the location of a maximum or minimum of a function, the tangent line has a slope of zero \(\displaystyle \left(0 \right )\).  We know the first derivative, \(\displaystyle f' (x)\), of a function \(\displaystyle f (x)\), gives us the value of the tangent line for a given value of \(\displaystyle x\).  Therefore, one needs to set the first derivative \(\displaystyle f' (x) = 0\).

First, we need to find the derivative of \(\displaystyle f(x)\), given below as 

\(\displaystyle f'(x)= \frac{2x+4}{\sqrt{x^2+4x-4}}\).  Plugging a value of \(\displaystyle x=2\) gives us

\(\displaystyle f'(2)= \frac{2(2)+4}{\sqrt{2^2+4(2)-4}}=\frac{8}{\sqrt{8}}=\frac{8}{\sqrt{2\cdot4}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\).

The correct answer is \(\displaystyle 0\)

Notice that this function is constant; it does not contain any variables. (\(\displaystyle a\) is stated to be a constant). Hence the derivative is \(\displaystyle 0\).

Looking at the term \(\displaystyle 6x^a\), since \(\displaystyle a\) is a constant, the derivative is \(\displaystyle a\times 6x^{a-1} = 6ax^{a-1}\).

The derivative of \(\displaystyle e^{\sin(x)}\) uses the Chain Rule. So the derivative is \(\displaystyle e^{\sin(x)} \times (\sin(x))' = \cos(x)e^\sin(x)\).

Remembering that \(\displaystyle a\) is just constant, we have the derivative of \(\displaystyle a\sin(ax)\) is \(\displaystyle a\cos(ax) \times a = a^2\cos(ax)\), also by the chain rule.

Adding our three results together gives

\(\displaystyle 6ax^{a-1}+\cos(x)e^{\sin(x)}+a^2\cos(ax)\).

It's important to take note that \(\displaystyle a\) is not a variable, it is a constant. Hence the derivative of \(\displaystyle a\) is \(\displaystyle 0\), not \(\displaystyle 1\). This idea will pop up in many places later on in Calculus 3.

Either the Product Rule, or the Constant Multiple Rule can be used to find the derivative of \(\displaystyle a\sin(t)\).

For example, the Product Rule will proceed as follows

\(\displaystyle (a\sin(t))' = (a)(\sin(t))'+(a)'(\sin(t))\)

\(\displaystyle = (a)(\cos(t))+(0)(\sin(t)) = a\cos(t)\).

Note that \(\displaystyle \frac{d}{dx} \tan \left( f(x) \right )= \sec ^2 \left(f(x) \right ) \cdot f'(x)\).

For this problem, \(\displaystyle f(x) = 3x^2 - 5x\Rightarrow f'(x)=6x - 5\).  Putting this together with the definition, we arrive at 

\(\displaystyle f' (x)= \sec ^2 \left(3x^2-5x \right ) \cdot \left(6x-5 \right )\).