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:

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

To find the derivative of the composite function, we must rewrite the composition as $\displaystyle f(g(x))$. Using the chain rule, the derivative of this function is equal to

$\displaystyle f'(g(x)) \cdot g'(x)$.

For our function,

$\displaystyle f'(x)=4x$ and $\displaystyle g'(x)=5$. These derivatives were found using the following rule:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$.

We now have all the pieces to use the chain rule formula, where we are evaluating the function at x=1:

$\displaystyle g(1)=5$

$\displaystyle g'(1)=5$

$\displaystyle f'(g(1))=f'(5)=20$

$\displaystyle 20 \cdot 5=100$

The derivative of the function is equal to

$\displaystyle f'(x)=\frac{e^t(4t^3-2t)-e^t(t^4+t^2)}{e^{2t}}=e^t(-t^4+4t^3+t^2-2t)$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

Because $\displaystyle \cos(x)$ and $\displaystyle \sec(x)$ are inverses of each other, we can simplify the given function to

$\displaystyle f(x)=e^{10x^2}$

Taking the derivative of this, we get

$\displaystyle f'(x)=20xe^{10x^2}$

by using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative: 

$\displaystyle d[sin(u)]=cos(u)du$

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function $\displaystyle f(x)=sin(10x)$ at the point $\displaystyle x=\pi$

The slope of the tangent is

$\displaystyle f'(x)=10cos(x)$

$\displaystyle f'(\pi)=10cos(\pi)=-10$

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative: 

$\displaystyle d[sin(u)]=cos(u)du$

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function $\displaystyle f(x)=sin(5\pi x^2)$ at the point $\displaystyle x=2$

The slope of the tangent is

$\displaystyle f'(x)=10\pi x cos(5\pi x^2)$

$\displaystyle f'(2)=10\pi (2) cos(5\pi (2)^2)=20\pi$

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative: 

$\displaystyle d[sin(u)]=cos(u)du$

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function $\displaystyle f(x)=sin^2(5x)$ at the point $\displaystyle x=\pi$

The slope of the tangent is

$\displaystyle f'(x)=10sin(5x)cos(5x)$

$\displaystyle f'(\pi)=10sin(5\pi)cos(5\pi)=0$