# Differentiating an integral

#### TsAmE

d/dx( ∫ 3x to x^3 (t^3 + 1)^10 dt )

I am not sure how to do this, except I know that you are suppose to use chain rule

#### Krizalid

MHF Hall of Honor
$$\displaystyle h(x)=\int_{f(x)}^{g(x)}{\delta (t)\,dt}\implies h'(x)=\delta \left( g(x) \right)g'(x)-\delta \left( f(x) \right)f'(x).$$

#### TsAmE

Makes sense, except I got to the part:

d/du( ∫ v to u (t^3 + 1)^10 dt )du/dx

but this doesnt make sense since my lower limit (v = 3x), isnt in terms of u (Wondering)

#### Random Variable

$$\displaystyle \frac{d}{dx} \int^{x^{3}}_{3x} (t^{3}+1)^{10} \ dt = \frac{d}{dx} \Big( F(x^{3}) - F(3x) \Big)$$ where F is an antiderivative of $$\displaystyle (t^3+1)^{10}$$

$$\displaystyle = F'(x^{3}) \ \frac{d}{dx} (x^{3}) - F'(3x) \ \frac{d}{dx} (3x) = \Big((x^{3})^3+1\Big)^{10} * 3x^{2} - \Big((3x)^{3}+1\Big)^{10} *3$$

$$\displaystyle = 3x^{2} (x^9+1)^{10} - 3(27x^{3}+1)^{10}$$

#### TsAmE

Thanks a lot. How would you do it in a form like this: d/du( ∫ v to u (t^3 + 1)^10 dt )du/dx, keeping the integral sign, and subbing in variables for the lower and upper limits?