To find the derivative of this expression, you must use the chain rule. This means you take the exponent of the binomial and multiply it by the coefficient in front of the binomial (1, in this case). Then, decrease the exponent of the binomial by 1. Lastly, find the derivative of the binomial.

Thus, your answer is:

 $\displaystyle 2(3x^2+9)(6x) or 12x(3x^2+9)$.

This problem involves the chain rule for derivatives. However, you must first rewrite the function as:

 $\displaystyle [(3x^3+4)^2]^{1/3}$ or $\displaystyle (3x^3+4)^{\frac{2}{3}}$.

Then, apply the chain rule (first multiply the exponent by the coefficient in front of the binomial [1], then decrease the exponent of the binomial by 1, and finally take the derivative of the binomial):

$\displaystyle \frac{2}{3}(3x^3+4)^{-1/3}(9x^2)$

When simplifiying, change negative exponents to positive ones. Therefore, the answer is:

$\displaystyle f{}'(x)=\frac{6x^2}{\sqrt[3]{3x^3+4}}$.

The correct answer is $\displaystyle x^{2}e^{x}(3+x)$.

We must use the product rule to solve. Remember that the derivative of $\displaystyle e^{x}$ is $\displaystyle e^{x}$.

$\displaystyle y'=(3x^{2})(e^{x})+(x^{3})(e^{x})$

$\displaystyle y'=x^{2}e^{x}(3+x)$

The derivative of $\displaystyle e^x$ is equal to $\displaystyle e^x$ therefore the first part of the equation remains the same.

The second part requires regular differential rules.

$\displaystyle x^ndx=nx^{n-1}$

Therefore when differentiating $\displaystyle x^e$ you get $\displaystyle ex^{e-1}$.

Combining the first and second part we get the final derivative:

$\displaystyle y'=e^x+ex^{e-1}$.

The derivative of $\displaystyle \ln(x)$ is $\displaystyle \frac{1}{x}$.

Therefore the integral of

$\displaystyle \int\frac{1}{x}dx= \ln(x)+C$ where C is some constant.

The rule for taking the derivative of $\displaystyle sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}$.

For this problem we need to remember to use the Chain Rule.

Since we are taking the derivative of,

 $\displaystyle sin^{-1}(4x)$ we need to take the derivative of the outside piece $\displaystyle sin^{-1}$ keeping the inside piece the same $\displaystyle 4x$, and then multiply the whole thing by the derivative of the inside piece $\displaystyle 4$.

Therefore the solution becomes:

$\displaystyle y'=\frac{1}{\sqrt{1-16x^2}}\cdot 4$,

$\displaystyle y'=\frac{4}{\sqrt{1-16x^2}}$.

The derivative of,

$\displaystyle \ln(1x)=\frac{1}{x}\cdot 1$.

The derivative of 

$\displaystyle \ln(2x)= \frac{1}{2x}\cdot2$

Therefore, using the Chain Rule the derivative of the function will become the derivative of the outside piece keeping the original inside piece. Then multiplying that by the derivative of the inside piece.

 $\displaystyle \ln(2x+6)= \frac{1}{2x+6}\cdot2$

  Recall the formal definition of the derivative:

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

When you evaluate this limit the output is f'(x). In this question f(x) = ln(x) so what this question is really saying is take the derivative of f(x) and evaluate it at 2.  

$\displaystyle f(x)=lnx$

Take the derivative:

$\displaystyle f'(x)=\frac{1}{x}$

Substitute the value 2 into the derivative:

$\displaystyle f'(2)=\frac{1}{2}$

This limit is very simple(almost too simple) because it asks for the limit at a location where there is no discontinuity. Fortunately, this makes taking the limit trivial.

Substitute x=4 into the function to evaluate the limit.

$\displaystyle f(4)=4^2-3=13$

As x becomes infinitely large, the x terms with the highest power dominate the function and the terms of lower order become negligible.  This means that near infinity, the x^5 term in the numerator and the 10X^5 term in the denominator are the only values necessary to evaluate the limit.

Simplify and evaluate:

$\displaystyle \frac{x^5}{10x^5}=\frac{1}{10}$