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$$e^y - e^x = xy^3$$

Find $\frac{dy}{dx}.$

I have tried logging both sides and have still not be able to find the derivative. I have additionally not been able to find the whole equations as a function of y. Please let me know the necessary steps I need to take in order to compute this

Thanks!

Siong Thye Goh
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Zacarura
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3 Answers3

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first substitute e^y=v and e^x=u then differentiate them and then separately find the derivatives of v and u.After this you just have to rearrange(and ofcourse you have to use implicit differentiation)

WhiteHole
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As user99163 commented, you need to use implicit differentiation.

So, consider the implicit function $$f=e^y - e^x - xy^3=0$$ $$f'_x=-e^x-y^3$$ $$f'_y=e^y-3xy^2$$ and by the implicit function theorem$$\frac{dy}{dx}=-\frac{f'_x}{f'_y}$$

You could not make $y$ explicit but, using Lambert function, you should get the awful

$$x=-\frac{e^y}{y^3}+W\left(\frac{1}{y^3}e^{\frac{e^y}{y^3}}\right)$$

Claude Leibovici
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  • Hey! Thanks so much, I was just wondering, to be sure, what the final answer of that differentiation was? – Zacarura Aug 19 '17 at 06:31
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$e^y y' - e^x = 1.y^3 + x. 3y^2. y'$

$e^y y' - e^x = y^3 + x. 3y^2. y'$

$e^y y' - x. 3y^2. y' = y^3 + e^x$

$(e^y - x. 3y^2)y' = y^3 + e^x$

$y' = \frac{y^3 + e^x}{e^y - x. 3y^2}$

Above $y'=\frac{dy}{dx}$

Raj Kumar Singh
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