Use the power rule to find the derivative.

$\displaystyle \frac{d}{dx}3x^4=12x^3$

$\displaystyle \frac{d}{dx}-2x^2=-4x$

$\displaystyle \frac{d}{dx}3x=3$

Thus, the derivative is $\displaystyle 12x^3-4x+3$.

Use the quotient rule to find the derivative.

$\displaystyle \frac{d}{dx}\frac{8}{x^3}=\frac{x^3(0)-8(3x^2)}{x^6}=\frac{-24x^2}{x^6}=\frac{-24}{x^4}$

Use the quotient rule to find the answer.

$\displaystyle \frac{7(2x)-x^2(0)}{49}$

Simplify.

$\displaystyle \frac{14x}{49}=\frac{2x}{7}$

Use the quotient rule to find the derivative.

$\displaystyle 2x^2\cos (x)+4x\sin (x)$

Use the power rule to find the derivative.

$\displaystyle \frac{d}{dx}-3x^2=-6$

Recall that the derivative of a constant is zero.

Thus, the derivative is -6.

To solve for $\displaystyle \frac{\mathrm{dy} }{\mathrm{d} x}$, we differentiate using normal rules, but when taking the derivative of y with respect to x, we must add the term we are solving for, $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$.

Taking the derivative, we get

$\displaystyle 2x+5e^xy+5e^x\frac{\mathrm{dy} }{\mathrm{d} x}=30y^2\frac{\mathrm{dy} }{\mathrm{d} x}$

using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

Finally, we solve for $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$:

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x+5e^xy}{30y^2-5e^x}$.

When looking at differentiability of piecewise functions over all $\displaystyle x$, first consider if the two functions are continuous and differentiable for all x. 

$\displaystyle \frac{1}{x}$ is discontinuous at $\displaystyle x=0$, but that's okay $\displaystyle \frac{1}{x}$ is restricted for $\displaystyle x< -1$. However $\displaystyle ln(-x)$ does not exist for $\displaystyle x\geq0$. Since this is the case, we can say that this piecewise function is not differentiable for all $\displaystyle x$.

To find the second derivative, you must first find the first derivative. When taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent. Therefore, the first derivative is: $\displaystyle 18x^2-4x$. Then, take the derivative again to get the second derivative: $\displaystyle f{}''(x)=36x-4$.

First, distribute the 9 to get $\displaystyle 36x^2-9$. Then, take the derivative, remembering to multiply the exponent by the coefficient in front of the x and then subtracting one from the exponent to get an answer of $\displaystyle f{}'(x)=72x$.

The derivative of the function is equal to

$\displaystyle f'(x)=3e^x+3xe^x-12\sin(x)$

and was found using the following rules:

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