To find the derivative, we must use the following rule:

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

Now, using the above rule, write out the derivative:

\(\displaystyle f'(t)=\frac{(50t^2-7t+1)(-4\sin(t)e^{\cos(t)})-4e^{\cos(t)}(100t-7)}{(50t^2-7t+1)^2}\)

The internal derivatives were found using the following rules:

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

Evaluated at \(\displaystyle t=0\), we get

\(\displaystyle f'(0)=\frac{-4(-7)}{1}=28\)

We will need to use the product rule and the chain rule to find the derivative.

\(\displaystyle y = 3xe^{\sin(x)}\). Start

\(\displaystyle y' = (3x)(e^{\sin(x)})' + (3x)'(e^{\sin(x)})\). Product Rule

\(\displaystyle y' = (3x)(\cos(x)e^{\sin(x)}) + (3)(e^{\sin(x)})\). Use the Chain Rule for \(\displaystyle (e^{\sin(x)})'\).

\(\displaystyle y' = 3x\cos(x)e^{\sin(x)} + 3e^{\sin(x)}\). Multiply

\(\displaystyle y' = 3e^{\sin(x)}(x\cos(x)+1)\). Factor out a \(\displaystyle 3\), and an \(\displaystyle e^{\sin(x)}\).

Although we could use the Product Rule to compute the derivative, it becomes much easier to find if we rewrite \(\displaystyle \tan(x) = \frac{\sin(x)}{\cos(x)}\).

\(\displaystyle y=50\cos(x)\tan(x)\). Start

\(\displaystyle y=50\cos(x)\frac{\sin(x)}{\cos(x)}\)

\(\displaystyle y = 50\sin(x)\)

\(\displaystyle y' = 50\cos(x)\).

Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have

\(\displaystyle x^y=yx\)

\(\displaystyle \ln(x^y)=\ln(xy)\)

\(\displaystyle y\ln(x)=\ln(x)+\ln(y)\)

\(\displaystyle y = \frac{\ln(y)}{\ln(x)}+1\)

\(\displaystyle y' = \frac{\ln(x)\frac{y'}{y}-\frac{\ln(y)}{x}}{\ln(x)^2}\). Quotient Rule + implicit differentiation.

\(\displaystyle y' = \frac{y'}{y\ln(x)}-\frac{\ln(y)}{x\ln(x)^2}\)

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

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

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

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

\(\displaystyle y'=\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}\)

The correct answer is \(\displaystyle \frac{x-(1+x^2)\tan^{-1}(x)}{x^2(1+x^2)}\).

Using the Quotient Rule and the fact \(\displaystyle \frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}\), we have

\(\displaystyle y = \frac{\tan^{-1}(x)}{x}\)

\(\displaystyle y' = \frac{x\frac{1}{1+x^2}-\tan^{-1}(x)}{x^2}\)

\(\displaystyle y' = \frac{\frac{x}{1+x^2}-\tan^{-1}(x)}{x^2}\)

\(\displaystyle y' = \frac{x}{x^2(1+x^2)}-\frac{\tan^{-1}(x)}{x^2}\)

\(\displaystyle y' = \frac{x}{x^2(1+x^2)}-\frac{\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}\)

\(\displaystyle y' = \frac{x-\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}\).

The Chain Rule is required here.

\(\displaystyle y=e^{x^{\frac{1}{2}}+x^2-\ln(x)}\). Start

The Chain Rule Proceeds as follows: \(\displaystyle f(g(x)) = f'(g(x))\times g'(x)\). In this case

 \(\displaystyle f(x)= e^x\)

and

 \(\displaystyle g(x) = \frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x}\).

Putting these into the Chain Rule, we get

\(\displaystyle y' = e^{x^\frac{1}{2}+x^2-\ln(x)}(\frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x})\)

Or the same to say

\(\displaystyle y' = (\frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x})e^{x^\frac{1}{2}+x^2-\ln(x)}\).

For the \(\displaystyle 6x^2\) term, we simply use the power rule to abtain \(\displaystyle 12x\). Since \(\displaystyle C\) is a constant (not a variable), we treat it as such. The derivative of any constant (or "stand-alone number") is \(\displaystyle 0\).

We use the Product Rule to find our answer here. The Product Rule formula is \(\displaystyle (fg)'(x) = f(x)g'(x)+f'(x)g(x)\).

Let \(\displaystyle f(x) = 7x^2\), \(\displaystyle g(x)= e^x\), then we have \(\displaystyle f'(x) = 14x\), \(\displaystyle g'(x)=e^x\).

Putting these into our formula, we have

\(\displaystyle y' = (7x^2)(e^x)+(e^x)(14x)\)

\(\displaystyle =e^x(7x^2+14x)\)

\(\displaystyle =7xe^x(x+2)\).

We can find the derivative of 

\(\displaystyle f(x)=5\sqrt{x}\)

using the power rule

\(\displaystyle \frac{d}{dx}x^a=ax^{a-1}\)

with \(\displaystyle a=\frac{1}{2}\)

so we have

\(\displaystyle f'(x)=5\frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{5}{2x^{1/2}}=\frac{5}{2\sqrt{x}}\)

The derivative of the displacement function is the velocity, so we need to find \(\displaystyle f'(t)\). We can use the power rule 

\(\displaystyle \frac{d}{dx} x^a=ax^{a-1}\)

with \(\displaystyle a=\frac{2}{5}\)

to get 

\(\displaystyle f'(t)=\pi\frac{2}{ 5}t^{-3/5}=\frac{2\pi}{ 5t^{3/5}}\)