Question

d/dx integral (0,x^5) sin^2t(dt)

Answer 1

To make sure I understand you correctly, is this your question?
\[
\frac{d}{dx}\int_0^{x^5}sin^2t\ dt
\]

Answer 2

0 and x^5 spots are switched

Answer 3

Okay, so \[\frac{d}{dx}\int_{x^5}^0sin^2t\ dt\]
Just apply the chain rule. Remember, \((f\circ g)'(x)=f'(g(x))\cdot g'(x)\). In this problem you'll have \(f(x)=\int_x^0 \sin^2t\ dt\), and \(g(x)=x^5\)

Answer 4

(-cos^2dt)(x^5)*5x^4???

Answer 5

Look @nbousca 's instruction closely!

Answer 6

f(x) = integral ....
-> f '(x) = derivative of integral

Answer 7

First we derive \(f\), which gives us \(f'(t)=-\sin^2t\) by the FTC. Then plug in \(g(x)\) for \(t\), then multiply by \(g'(x)\).