$$\int_0^y e^{-t^2} dt + \int_0^{x^2} \sin^2(t) dt = 0$$ $$\dfrac{dy}{dx} = \ ?$$ I've been learning Integrals and came across this problem, would love a detailed explanation on how to solve this.
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It's not $u$ it is $x$ I have a bad hand writing. – ATheCoder Dec 24 '16 at 09:10
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Thanks for the edit Patrick really appreciated <3 – ATheCoder Dec 24 '16 at 09:11
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If you search in Google 14.1 Second fundamental theorem of calculus, the first entry is for simple examples from a page of MIT. Then you need read and understand these, with your summands. When you take the derivative of $y=y(x)$ you need write $y'=y'(x)$. Good luck. – Dec 24 '16 at 09:19
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Differentiate with respect to $x$ using the Leibniz rule:
$$e^{-y^2} \dfrac{dy}{dx} + 2x\sin^2(x^2) = 0$$
(Though it is Christmas Eve, a day on which I am more than usually liable to make algebraic errors.)
Patrick Stevens
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1Should that be $e^{-y^2} \dfrac{dy}{dx} + 2x \cdot \sin^2(x^2) = 0$? – JimmyK4542 Dec 24 '16 at 09:17
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Differentiating the integral we have using Leibnitz rule like this question, we get, $$e^{-y^2} \frac{dy}{dt} - 0 + 2x\frac{dx}{dt} \sin^2 x^2 - 0 = 0$$ Thus on rearranging, we have, $$e^{-y^2} \frac{dy}{dt} = -2x\frac{dx}{dt} \sin^{2}x^2 \Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2x\sin^{2} x^2}{e^{-y^2}}$$ Hope it helps.
