Why is D the correct option here? Shouldn't we just replace t^2 with (2x)^2? Why that 2 appears behind the sin?
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$\displaystyle \dfrac{d}{dx} \bigg[\int_a^u f(t) \, dt \bigg] = f(u) \cdot \color{red}{\dfrac{du}{dx}}$ $\displaystyle \dfrac{d}{dx} \bigg[\int_0^{2x} \sin(t^2) \, dt \bigg] = \sin[(2x)^2] \cdot \color{red}{2}$
Leibniz's rule: . Here so , so , and so
Last edited by HallsofIvy; Nov 26th 2017 at 11:16 AM.