Use the product rule to find the derivative of the function. 

$\displaystyle f(x)=(5x-3)(2x+1)$

$\displaystyle f'(x)= (5x-3)*2+(2x+1)*5$

$\displaystyle f'(x)= 10x-6+10x+5=20x-1$

The derivative of any function of e to any exponent is equal to the function multiplied by the derivative of the exponent. 

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

$\displaystyle y=ln(x^2)-ln(x^2+1)$

$\displaystyle y'= \frac{1}{x^2}*2x-\frac{1}{x^2+1}*2x$

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

Factoring out an x gives you $\displaystyle \frac{2}{x^3+x}$.

$\displaystyle y''=2(x^3+x)^{-1}=-2(x^3+x)^{-2}*(3x^2+1)=\frac{-2(3x^2+1)}{(x^3+x)^2}$

Need to use the power rule which states: $\displaystyle \frac{d}{dx}u^n=nu^{n-1}\frac{du}{dx}$

In our problem $\displaystyle \frac{du}{dx}=-3sin(3x)$

For $\displaystyle f(x) = x^{n}$, the nth derivative is $\displaystyle N!$.  As an example, consider $\displaystyle f(x) = x^{3}$. The first derivative is $\displaystyle 3x^{2}$, the second derivative is $\displaystyle 3*2x\: (or\: 6x)$, and the third derivative is $\displaystyle 3*2*1,\: or \: 3!$.  For the question being asked, the 99th derivative of $\displaystyle x^{99}$ would be $\displaystyle 99!$.  The 66th derivative of $\displaystyle x^{66}$ would be $\displaystyle 66!$, and any higher derivative would be zero, since the derivative of any constant is zero.  Thus, for the given function, the 99th derivative is $\displaystyle 99!$.

$\displaystyle f(\pi ) = 4\pi^{2} - sin(\pi ) - 10\pi > 0$

$\displaystyle f'(\pi ) = 8\pi - cos(\pi) -10 > 0$, meaning $\displaystyle f(x)$ is increasing when $\displaystyle x =$$\displaystyle \pi$.

$\displaystyle f''(\pi ) = 8 + sin(\pi ) > 0$, meaning $\displaystyle f(x)$ is concave up when $\displaystyle x =$$\displaystyle \pi$.

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:

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