$\displaystyle \int \frac{e^{x^{\frac{1}{2}}}}{x^{\frac{1}{2}}}dx$

We rewrite the denominator as a negative exponenet in the numerator to make the u-substitution easier to see:

$\displaystyle \int \frac{e^{x^{\frac{1}{2}}}}{x^{\frac{1}{2}}}dx = \int e^{x^{\frac{1}{2}}}}{x^{- \frac{1}{2}}dx$

$\displaystyle Let\: u=x^{1/2}, du=\frac{1}{2}x^{-1/2}dx.$

$\displaystyle \int e^{x^{\frac{1}{2}}}}{x^{- \frac{1}{2}}dx = 2\int \frac{1}{2}e^{x^{\frac{1}{2}}}}{x^{- \frac{1}{2}}dx = 2 \int e^udu = 2e^u + c = 2e^{x^{\frac{1}{2}}} + c$, which is our final answer.

$\displaystyle \int 2sinx*cosxdx$

$\displaystyle Let\: u=sinx,\: du=cosxdx.$

 $\displaystyle \int 2sinx*cosxdx = \int 2udu = u^2 +c = sin^2x +c$, which is our final answer.

$\displaystyle \int \frac{e^{(1+lnx)}}{x}dx$

$\displaystyle Let \ u = 1+ln(x)), \ du=\frac{1}{x}dx.$

 $\displaystyle \int \frac{e^{(1+lnx)}}{x}dx = \int e^udu= e^u +c = e^{(1+lnx)}+c$, which is our final answer.

The derivative of the difference of two functions is the difference of the derivative of the two functions:
$\displaystyle \frac{dy}{dx} = 2x - \frac{d}{dx}(sin(x))$
      $\displaystyle = 2x - cos(x)$

We can write the function as

 $\displaystyle \frac{1}{(3x^2+5)^4}=(3x^2+5)^{-4}}$.  

Let  $\displaystyle u(x)=3x^2+5\Rightarrow \frac{du}{dx}=6x$.  

We then have 

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}=-4(3x^2+5)^{-5}*(6x)=\frac{-24x}{(3x^2+5)^5}$.

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

Using the chain rule, $\displaystyle if \ f(x)=g(h(x))$, $\displaystyle then\ f'(x)=g'(h(x))*h'(x)$,

we observe the following:

$\displaystyle g(x)= e^x \ so\ g'(x)= e^x$.

$\displaystyle h(x) = x^2 \ so \ h'(x)=2x$.

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

$\displaystyle f'(x)=g'(h(x))*h'(x) = e^{x^{2}}*2x = 2x*e^{x^{2}}$, which is our final answer.

$\displaystyle f(x)=\frac{sin(x)}{x}$

We evaluate this derivative using the quotient rule:

$\displaystyle if \ f(x)=\frac{g(x)}{h(x)}$,

$\displaystyle then\ f'(x)=\frac{g'(x)*h(x) - g(x)*h'(x)}{h(x)^{2}}$.

Apply the above formula:

$\displaystyle f'(x)=\frac{cos(x)*x - sin(x)*1}{x^2}$

$\displaystyle f'(x)=\frac{x*cos(x) - sin(x)}{x^2}$, which is our final answer.

First we must find the first derivative of f(x).

f'(x) = 4x³ + 12x^–5

To find the slope of the tangent line of f(x) at 5, we merely have to evaluate f'(x) at 5:

f'(5) = 4*5³ + 12* 5^–5 = 500 + 12/3125 = 500.00384

$\displaystyle g{}'(x)=tan^{-1}(4x^3+7x^2)$

using the identity:

 $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\ tan^{-1}[(f(x)]=\frac{f{}'(x)}{1+[f(x)]^2}$

$\displaystyle g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

The Quotient Rule applies when differentiating quotients of functions.  Here, $\displaystyle \small f(x)$ equals the quotient of two functions, $\displaystyle \small x^2$ and $\displaystyle \small 2x+7$.  Let $\displaystyle \small l(x)=2x+7$ and $\displaystyle \small h(x)=x^2$.  (Think: $\displaystyle \small l(x)$ is the "low" function or denominator and $\displaystyle \small h(x)$ is the "high" function or numerator.)  The Quotient Rule tells us to multiply the "low" function by the derivative of the "high" function, subtract the product of the "high" function and the derivative of the "low" function, and then divide the result by the square of the "low" function.  In other words,

$\displaystyle \small \small f'(x)=\frac{l(x)h'(x)-h(x)l'(x)}{[l(x)]^2}$

Here, $\displaystyle \small l(x)=2x+7$ so $\displaystyle \small l'(x)=2$.  Similarly, $\displaystyle \small h(x)=x^2$ so $\displaystyle \small h'(x)=2x$.

Then

$\displaystyle \small f'(x)=\frac{(2x+7)(2x)-x^2(2)}{(2x+7)^2}$

$\displaystyle \small f'(x)=\frac{(4x^2+14x)-2x^2}{(2x+7)^2}$

$\displaystyle \small f'(x)=\frac{2x^2+14x}{(2x+7)^2}$

Factoring out $\displaystyle \small 2x$ from the numerator gives

$\displaystyle \small \small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

$\displaystyle \small f'(x)=\frac{-2x(x+7)}{(2x+7)^2}$ inverts the order of the numerator, subtracting $\displaystyle \small l(x)h'(x)$ from $\displaystyle \small h(x)l'(x)$.

$\displaystyle \small f'(x)=\frac{2x(3x+7)}{(2x+7)^2}$ adds the products in the numerator, rather than subtracting them.

$\displaystyle \small f'(x)=\frac{2x(x+7)}{(2x+7)}$ fails to square the denominator.