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I have recently posted about solving a much easier problem using the theorem in question. Now, I decided to try and tackle a much more tricky problem. Namely:

$$f(x) = \frac{d}{dx} \int_{\sin(x)}^{2x}\cos(t) dt$$ Find $\frac{df}{dx}$

This is my solution:
1. Rewrite this integral as a sum of two more-easily-computable ones: $$f(x) = \frac{d}{dx}\int_{\sin(x)}^0 \cos(t) dt + \frac{d}{dx} \int_0^{2x}\cos(t)dt$$ 2. Evaluate the first integral: $$f'(\sin(x)) = \cos(x) \frac {df}{d\sin(x)}$$ $$\int_{\sin(x)}^0 \cos(t) dt = \cos(x)\cos(\sin(x))$$ 3. Evaluate the second integral: $$\int_0^{2x}\cos(t)dt = 2\cos(2x)$$ 4. Find the derivative: $$ \frac{d}{dx}\left(\cos(x)\cos(\sin(x))+ 2\cos(2x) \right)= \dots$$ The computation of this derivative is not the kernel of my problem and so I can skip it here.
Do you think that my method / solution is correct?

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You have done every thing right except for $$ \frac {d}{dx} \int_{sin(x)}^0 \cos(t) dt = \cos(x)\cos(\sin(x))$$ where you missed a negative sign.

The correct answer should be $$\frac {d}{dx} \int_{sin(x)}^0 \cos(t) dt = -\cos(x)\cos(\sin(x))$$

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    $\begingroup$ Is this because of this fact that when we apply the Fundamental Theorem of Calculus, the term containing the variable has to be the upper bound? $\endgroup$ – Aemilius Jan 18 '18 at 14:57
  • $\begingroup$ @Aemilius Perfect! It is exactly because of the Fundamental Theorem of Calculus. $\endgroup$ – Mohammad Riazi-Kermani Jan 18 '18 at 15:03
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your integral is given by $$\sin(2x)-\sin(\sin(x))$$ and the first derivative is $$2 \cos (2 x)-\cos (x) \cos (\sin (x))$$

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    $\begingroup$ You simply computed this integral - the problem is required to be solved by the fundamental theorem. $\endgroup$ – Aemilius Jan 18 '18 at 14:53

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