Fundamental Theorem of Calculus With Weird Limits $$f(x) =\int_6^{x^3} \sin^3(5t) \, dt$$
what is $f'(x)$?
I know I have to use fundamental theorem of calculus, but what do I do about the $x^3$ and $t^3$?
 A: The fundamental theorem of calculus:
if $F(x)=\int_{a}^{x}f(t)dt$ then $F'(x_0)=f(x_0)$.
We have $F(x)=\int_{6}^{x^3}\sin ^3 (5t)dt$.
Let $\alpha = x^3$. then $F(\alpha ^ {\frac{1}{3}})=\int_{6}^{\alpha}\sin ^3 (5t)dt$
Now derive: $(F(\alpha ^ {\frac{1}{3}}) )'=\frac{1}{3}\alpha ^ {\frac{-2}{3}}F'(\alpha ^{\frac{1}{3}}) $
That was from chain rule. But we also know from FTC that it is equal to $\sin ^3(5 \alpha)$.
So you have $\frac{1}{3}\alpha ^ {\frac{-2}{3}}F'(\alpha ^{\frac{1}{3}}) =\sin ^3(5 \alpha)$. Solve for $F'(\alpha ^ {\frac{1}{3}})$ and return to $x$ rather than $\alpha$.
Also, next time do try to show more effort or indication that you tried something.
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A: Use the chain rule:
\begin{align}
y & = \int_6^{x^3} \sin^3(5t)\,dt \\[12pt]
y & = \int_6^u \sin^3(5t)\,dt \\[6pt]
u & = x^3 \\[10pt]
\text{Therefore } \frac{dy}{du} & = \sin^3(5u) \\[6pt]
\text{and } \frac{du}{dx} & = 3x^2 \\[10pt]
\text{so } & \underbrace{\frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx}}_\text{chain rule} = \sin^3(5u)\cdot 3x^2 \\[15pt]
& = \sin^3(5x^3)\cdot 3x^2.
\end{align}
