Differentiate: 

$\displaystyle f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$\displaystyle f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$  

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text. 

Let,

$\displaystyle u= x^2+1$

Therefore, 

$\displaystyle \frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as: 

$\displaystyle f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

 $\displaystyle f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to $\displaystyle u$ in the second term using the Second FTOC and then convert back to $\displaystyle x$. 

$\displaystyle \frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$ 

Therefore we have, 

$\displaystyle f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$\displaystyle f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

Use the power rule to find the derivative.

$\displaystyle \frac{d}{dx}12x^3=36x^2$

To take the derivative you need to bring the power down to the front of the equation, multiplying it by the coefficient and then drop the power.

So:

$\displaystyle f(x)=3x+2$  becomes $\displaystyle f'(x)=3$

because the degree of "x" is just one, and once you multiply 3 by 1 you get 3 and drop the power of "x" to 0. The second term is just a constant and the derivative of any term is just 0. 

To do this problem you need to use the quotient rule. So you do

(low)(derivative of the high)-(high)(derivative of the low) all divided by the bottom term squared.

So:

$\displaystyle \frac{x(2)-2x(1))}{x^2}$

Which, when simplified is:

$\displaystyle \frac{2x-2x}{x^2}=0$

This is the product rule, which is: (derivative of the first)(second)+(derivative of the second)(first)

So:

$\displaystyle (9x^2)(x)+(1)(3x^3)=12x^3$

This is a combination of chain rule and quotient rule.

So:

$\displaystyle 2x(\frac{1}{x})+\frac{0-1}{x^2}(x^2)=2-\frac{x^2}{x^2}$

Which when simplified you get:

$\displaystyle f'(x)=2-1=1$

This problem is just addition of derivatives using trigonometric functions.

So:

$\displaystyle f'(x)=cos(x)+sec^2(x)$

The is a quotient rule using a trigonometric function.

So:

$\displaystyle f'(x)=\frac{x^2(sec^2(x))-tan(x)(2x)}{x^4}$

You can pull out an "x" and cancel it to get:

$\displaystyle f'(x)=\frac{xsec^2(x)-2tan(x)}{x^3}$

This is the same concept as a normal derivative just with a negative in the exponent.

$\displaystyle f'(x)=-10(x^{-9})$ 

which becomes:

$\displaystyle f'(x)=\frac{-10}{x^9}$

This is a power rule that can utilize u-substitution.

So  $\displaystyle \small f'(x)=(u)^2*u'$

where $\displaystyle \small u=x^2$

So you get:

$\displaystyle \small f'(x)=2u*u'$

Plug "u" back in and you get:

$\displaystyle \small f'(x)=2(x^2)(2x)$