\(\displaystyle \int \frac{ln(x)}{x}dx\)

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

\(\displaystyle \int \frac{ln(x)}{x}dx = \int udu = \frac{1}{2}u^2 + c = \frac{1}{2}ln(x)^2 +c\), which is our final answer.

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

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:

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

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