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By FTC, if the upper bound in the integral was simply $x$ then it would simply be $f'(x)=\sin \left(\pi t^{2}\right)$ right? But the upper bound is $x^3$, so how can I find the derivative?

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    $\begingroup$ Take $g(y)=\int_{-2020}^y\sin(\pi t^2)dt$ and then, as $f(x)=g(x^3)$, calculate the derivative of $f$ using the chain rule. $\endgroup$ – Stinking Bishop Jan 21 at 18:11
  • $\begingroup$ A better version of the FTC is $$\mathrm{\partial}_x\left[\int_{a(x)}^{b(x)}f(s)\mathrm{d}s\right]=f(b(x))b'(x)-f(a(x))a'(x)$$ $\endgroup$ – K.defaoite Jan 22 at 3:09
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Hint : $$f = g \circ h$$

where $$g(x)=\int_{-2020}^{x} \sin \left(\pi t^{2}\right)dt \quad \text{and} \quad h(x)=x^3$$

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