Differentiating an integral

Apr 2010
58
0
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
 
Apr 2010
58
0
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)
 
May 2009
959
362
\(\displaystyle \frac{d}{dx} \int^{x^{3}}_{3x} (t^{3}+1)^{10} \ dt = \frac{d}{dx} \Big( F(x^{3}) - F(3x) \Big) \) where F is an antiderivative of \(\displaystyle (t^3+1)^{10} \)

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

\(\displaystyle = 3x^{2} (x^9+1)^{10} - 3(27x^{3}+1)^{10} \)
 
Apr 2010
58
0
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?