To find the second derivative, first we need to find the first derivative. The derivative of a natural log is the derivative of operand times the inverse of the operand. So for the given function, we get the first derivative to be 

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

Now, we have to take the derivative of the first derivative. To simplify this, we can rewrite the function to be \(\displaystyle f'(x)=2(2x-1)^-^1\). From here we can use the chain rule to solve for the derivative. First, multiply by the exponent and find the new exponent by subtracting the old one by one. Next multiply by the derivative of (2x-1) and then simplify. Thus, we get 

\(\displaystyle f''(x)=2\cdot-1\cdot2(2x-1)^-^2=-\frac{4}{(2x-1)^2}\).

To find the second derivative, first we need to find the first derivative. So for the given function, we get the first derivative to be 

\(\displaystyle f'(x)=tan(x)sec(x)\). 

Now we have to take the derivative of the derivative. To do this we need to use the product rule as shown below

 Thus, we get

 \(\displaystyle \\f''(x)=tan(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}sec(x)+sec(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}tan(x)\\f''(x)=tan(x)\cdot sec(x)\cdot tan(x)+sec(x)\cdot sec^2(x)\\f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))\).

To find the second derivative, first we need to find the first derivative. To find the first derivative we need to use the quotient rule as follows. So for the given function, we get the first derivative to be 
\(\displaystyle \\f'(x)=\frac{x\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2x+1)-(x^2+2x+1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x)}{x^2}\\f'(x)=\frac{x\cdot (2x+2)-(x^2+2x+1)\cdot 1}{x^2}\\f'(x)=\frac{x^2-1}{x^2}\). 

Now we have to take the derivative of the derivative. To do this we need to use the quotient rule as shown below.

 Thus, we get 

\(\displaystyle \\f''(x)=\frac{x^2\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2-1)-(x^2-1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2)}{x^4}\\ f''(x)=\frac{2x^3-2x^3+2x}{x^4}\\f''(x)=\frac{2x}{x^4}=\frac{2}{x^3}\)

There are two seprate functions that make up \(\displaystyle f(x)\). There is \(\displaystyle 2sin(2x)\) and \(\displaystyle e^x\).

On a general note, 

\(\displaystyle \frac{d}{dx}sin(ax)=acos(ax)\) and

\(\displaystyle \frac{d}{dx}cos(ax)=-asin(ax)\).

Also, 

\(\displaystyle \frac{d}{dx}e^u=e^u \cdot u'\)

With that said, let's calculate \(\displaystyle f'(x)\):

\(\displaystyle f'(x)= 4cos(2x)+e^x\). Notice the \(\displaystyle e^x\) term is still unchanged.

And now let's calculate \(\displaystyle f''(x)\).

\(\displaystyle f'(x)= -8sin(2x)+e^x\)

To find a and b, first let's calculate \(\displaystyle f'(x)\).

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

Remember that \(\displaystyle \frac{d}{dx}e^{ax}=ae^{ax}\), where a is real number.

\(\displaystyle a\) is simply the coefficient in front of the exponential, which simplifies to 1, and \(\displaystyle b\) is the power of the exponent, which is 2.

In order so solve the derivative with the respect to x, implicit differentiation is required.  The notation for finding the derivative of the function with the respect to x is:

\(\displaystyle \frac{dy}{dx}\)

Take the derivative and apply chain rule where necessary.

\(\displaystyle y^2=x^2+5y\)

\(\displaystyle 2y \cdot \frac{dy}{dx}=2x +5 \cdot \frac{dy}{dx}\)

\(\displaystyle 2y \cdot \frac{dy}{dx}-5 \cdot \frac{dy}{dx}=2x\)

\(\displaystyle \frac{dy}{dx}[2y-5]=2x\)

\(\displaystyle \frac{dy}{dx}=\frac{2x}{2y-5}\)

To solve this derivative, we need to use logarithmic differentiation. This allows us to use the logarithm rule \(\displaystyle \small lna^{b}=blna\) to solve an easier derivative.

Let \(\displaystyle \small y=lnx^{lnx}\).

Now we'll take the natural log of both sides to get 

\(\displaystyle \small lny=ln(lnx^{lnx})=lnxln(lnx)\).

Now we can use implicit differentiation to solve for \(\displaystyle \small y'\).

The derivative of \(\displaystyle \small lny\) is \(\displaystyle \small y'/y\), and the derivative of \(\displaystyle \small \small \small lnx(ln(lnx))\) can be found using the product rule, which states 

\(\displaystyle \small \frac{\mathrm{d} }{\mathrm{d} x}(u\cdot v)=u'v+uv'\) where \(\displaystyle \small u\) and \(\displaystyle \small v\) are functions of \(\displaystyle \small x\).

Letting \(\displaystyle \small u=lnx\) and \(\displaystyle \small v=ln(lnx)\) 

(which means \(\displaystyle \small u'=1/x\) and \(\displaystyle \small v'=1/(xlnx)\)) we get our derivative to be \(\displaystyle \small ln(lnx)/x+1/x\).

Now we have \(\displaystyle \small y'/y=ln(lnx)/x+1/x\), but \(\displaystyle \small y=lnx^{lnx}\), so subbing that in we get 

\(\displaystyle \small \small y'/(lnx^{lnx})=ln(lnx)/x+1/x\).

Multiplying both sides by \(\displaystyle \small lnx^{lnx}\), we get 

\(\displaystyle \small y'=lnx^{lnx}(ln(lnx)/x+1/x)\).

That is our derivative.

The derivative of the function is equal to

\(\displaystyle -3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}\)

and was found using the following rules:

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

The first derivative of the function is equal to

\(\displaystyle f'(x)=2e^{3x}+6xe^{3x}+\sec^2(x)\)

The second derivative of the function (the derivative of the above function) is

\(\displaystyle f''(x)=12e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)\)

The following rules were used for the derivatives:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)\)

The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

\(\displaystyle -\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)\)

The derivative was found using the following rules:

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