Evaluating the limit directly will produce an indeterminant solution of \(\displaystyle \small \tiny{\frac{0}{0}}\).

The limit definition of a derivative is \(\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)\). However, the alternative form, \(\displaystyle \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(x)\), better suits the given limit.

Let \(\displaystyle f(x)=\sin^2x\) and notice \(\displaystyle f\left(-\frac{\pi}{2}\right)=1\). It follows that \(\displaystyle {\lim_{x \rightarrow -\frac{\pi}{2}} \frac{\sin^2-1}{x-\left(-\frac{\pi}{2} \right )} = \frac{d}{dx}\sin^2x}\).  

Thus, the limit is \(\displaystyle {2\sin\left(-\frac{\pi}{2} \right ) \cos\left(-\frac{\pi}{2}\right)} = 0.\)

Evaluating the derivative directly will produce an indeterminant solution of \(\displaystyle \small \tiny{\frac{0}{0}}\).

The limit definition of a derivative is \(\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)\). However, the alternative form, \(\displaystyle \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(x)\), better suits the given limit.

Let \(\displaystyle f(x)=x^{49}\) and notice \(\displaystyle f(1)=1\). It follows that \(\displaystyle {\lim_{x \rightarrow 1} \frac{x^{49}-1}{x-1} = \frac{d}{dx}x^{49}}\).  Thus, the limit is \(\displaystyle 49(1)^{48}=49\).

The correct answer is 11.

Taking the derivative of \(\displaystyle h(x)\) involves the product rule, and the chain rule.

\(\displaystyle h'(x)=[f(x+2)][-g'(-x)]+[g(-x)][f'(x+2)]\)

\(\displaystyle = -g'(-x)f(x+2)+g(-x)f'(x+2)\)

Substituting \(\displaystyle x=1\) into both sides of the derivative we get

\(\displaystyle h'(1)= -g'(-1)f(3)+g(-1)f'(3) = -1\cdot1 + 3 \cdot 4 = 11\).

If we recall the definition of a derivative of a function \(\displaystyle \small f(x)\) at a point \(\displaystyle \small x\), one of the definitions is 

\(\displaystyle \small f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\).

If we compare this definition to the limit 

\(\displaystyle \small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}\)

we see that that this is the limit definition of a derivative, so we need to find the function \(\displaystyle \small f(x)\) and the point \(\displaystyle \small x\) at which we are evaluating the derivative at. It is easy to see that the function is \(\displaystyle \small f(x)=\tan x\) and the point is \(\displaystyle \small x=0\). So finding the limit above is equivalent to finding \(\displaystyle \small f'(0)\).

We know that the derivative is \(\displaystyle \small f'(x)=\sec^2(x)\), so we have

\(\displaystyle \small \small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}=f'(0)=\sec^2(0)=1^2=1\).

Write the definition of the limit.

\(\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\)

Substitute \(\displaystyle x=2\).

\(\displaystyle \lim_{h\to 0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0}\frac{\sqrt{2+h}-\sqrt2}{h}\)

Since \(\displaystyle h\) is approaching to zero, it would be best to evaluate when we assume that \(\displaystyle h\) is progressively decreasing.  Let's assume \(\displaystyle h=0.1, 0.0001, 0.000001\) and check the pattern.

\(\displaystyle \frac{\sqrt{2+0.1}-\sqrt2}{0.1}=0.349241\)

\(\displaystyle \frac{\sqrt{2+0.0001}-\sqrt2}{0.0001}=0.353549\)

\(\displaystyle \frac{\sqrt{2+0.000001}-\sqrt2}{0.000001}=0.353553\)

The best answer is:  \(\displaystyle 0.35355\)

Computation of the derivative requires the use of the Product Rule and Chain Rule. 

The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:

\(\displaystyle f(x)*g(x)\)

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)*g'(x)+g(x)*f(x)\)

This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."

In the problem statement, we are given:

\(\displaystyle f(x)=3x^2(x^2-5x+2)^5\)

\(\displaystyle 3x^2\) is the "First" function, and \(\displaystyle (x^2-5x+2)^5\) is the "Second" function. 

The "Second" function requires use of the Chain Rule. 

When:

\(\displaystyle y=(f(x))^a \rightarrow \frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)\)

Applying these formulas results in:

\(\displaystyle f'(x)=(3x^2)[5(x^2-5x+2)^4(2x-5)]+(x^2-5x+2)^5(6x)\)

Simplifying the terms inside the brackets results in:

\(\displaystyle f'(x)=(3x^2)[(10x-25)(x^2-5x+2)^4]+(x^2-5x+2)^5(6x)\)

We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".

\(\displaystyle f'(x)=(x^2-5x+2)^4[(3x^2)(10x-25)+(6x)(x^2-5x+2)]\)

Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:

\(\displaystyle f'(x)=(x^2-5x+2)^4[30x^3-75x^2+6x^3-30x^2+12x]\)

Simplifying this results in one of the answer choices:

\(\displaystyle f'(x)=(x^2-5x+2)^4(36x^3-105x^2+12x)\)

Recall that one definition for the derivative of a function \(\displaystyle f(x)\ at\ x=a\) is \(\displaystyle f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\).

This means that this question is asking us to find the value of the derivative of \(\displaystyle sin(x)\) at \(\displaystyle x=\frac{\pi}{6}\).

Since 

\(\displaystyle \frac{d}{dx}sin(x)=cos(x)\) and \(\displaystyle cos(\frac{\pi}{6})=\frac{\sqrt3}{2}\), the value of the limit is \(\displaystyle \frac{\sqrt3}{2}\).

Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of \(\displaystyle \tan^{-1}(x)\).

Product Rule:

 \(\displaystyle \frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)\)

\(\displaystyle \frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}\)

Applying these two rules results in:

\(\displaystyle \frac{dy}{dx}=\frac{d}{dx}[x^2\tan^{-1}(3x^5)]\)

\(\displaystyle \frac{dy}{dx}=(x^2)(\frac{1}{1+(3x^5)^2})(15x^4)+\tan^{-1}(3x^5)(2x)\)

\(\displaystyle \frac{dy}{dx}=\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)\)

This matches one of the answer choices.

In order to find \(\displaystyle f'(x)\), we need to remember how to find \(\displaystyle f'(x)\) by using the definition of derivative.

Definition of Derivative:

\(\displaystyle f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}\)

Now lets apply this to our problem.

\(\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{(x+h)^3-x^3}{h}\)

Now lets expand the numerator.

 \(\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}\)

We can simplify this to

\(\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{3x^2h+3xh^2+h^3}{h}\)

Now factor out an h to get

\(\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{h(3x^2+3xh+h^2)}{h}\)

We can simplify and then evaluate the limit.

\(\displaystyle f'(x)=\lim_{h\rightarrow0} 3x^2+3xh+h^2\)

\(\displaystyle \\ \\ =3x^2+3x(0)+0^2 \\ =3x^2\)

In order to find \(\displaystyle f'(x)\), we need to remember how to find \(\displaystyle f'(x)\) by using the definition of derivative.

Definition of Derivative:

\(\displaystyle f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}\)

Now lets apply this to our problem.

\(\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{10(x+h)^2-10x^2}{h}\)

Now lets expand the numerator.

 \(\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{10x^2+20xh+10h^2-10x^2}{h}\)

We can simplify this to

\(\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{20xh+10h^2}{h}\)

Now factor out an h to get

\(\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{h(20x+10h)}{h}\)

We can simplify and then evaluate the limit.

\(\displaystyle f'(x)=\lim_{h\rightarrow0} 20x+10h\)

\(\displaystyle \\ \\ =20x+10(0) \\ =20x\)