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I am trying to find the derivative $\frac{dy}{dx}$ of the following function $ye^{xy} = \sin x$ using implicit differentiation.


What I have is the following using product rule:

$$\frac{d}{dx}ye^{xy}=\frac{d}{dx}\sin x$$ $$\implies y^2e^{xy} + \frac{dy}{dx}e^{xy}=\cos x$$

However, I noticed in the solution manual it has it as the following and not sure why:

$$\frac{d}{dx}ye^{xy}=\frac{d}{dx}\sin x$$ $$\implies ye^{xy} \left(x\frac{dy}{dx} + y\right) + \frac{dy}{dx}e^{xy}=\cos x$$

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  • $\begingroup$ $y$ is a function of $x$, right? So is $\frac{\mathrm{d}}{\mathrm{d}x} y = y$? $\endgroup$ Mar 18, 2020 at 4:07

3 Answers 3

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Your issue is that $y\frac{d}{dx}e^{xy}=ye^{xy}[\frac{d}{dx}xy]=ye^{xy}[y+xy'].$

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With $y$ as a function of $x$, we have $\dfrac{\operatorname d}{\operatorname dx}e^{xy}=e^{xy}(y+x\dfrac{\operatorname dy}{\operatorname dx})$.

The book is right.

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$$\frac{d}{dx}ye^{xy}=\frac{d}{dx}\sin x$$ This line is not correct: $$\implies \color {red}{y^2e^{xy}} + \frac{dy}{dx}e^{xy}=\cos x$$

Note that you can do this: $$\frac{d}{dx}ye^{xy}=e^{xy}\frac{dy}{dx}+y\frac{de^{xy}}{dx}$$ And apply the chain rule: $$\frac{de^{xy}}{dx}=\frac{de^{xy}}{dxy}\frac{dxy}{dx}=e^{xy}(y'x+y)$$ Therefore: $$\frac{d}{dx}ye^{xy}=e^{xy}y'+ye^{xy}(y'x+y)$$

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