I'm trying to compute the (implicit) derivative $\frac{dy}{dx}$ from the function


If I use $\ln$ on both sides I can isolate $y$ and find the derivative:

$$\ln(e^{2y}) = \ln(x^3)$$ $$2y = 3\ln(x)$$ $$y=\frac{3}{2} \ln(x)$$ $$y' = \frac{3}{2x}$$

But if I use implicit differentiation I get:

$$\frac{d}{dx}(e^{2y}) = \frac{d}{dx}x^3$$ $$\frac{d}{dy}(e^{2y})\frac{d}{dx}(y) = 3x^2$$ $$2e^{2y}\cdot y'=3x^2$$ $$y'=\frac{3x^2}{2e^{2y}}$$

I know both methods should give the same result. What am I missing?


  • 3
    $\begingroup$ Both methods give the same result. What you missed is that $e^{2y} = x^3. \qquad$ $\endgroup$ – Michael Hardy Sep 14 '17 at 18:13
  • 10
    $\begingroup$ Maybe another thing you're missing is this: You can't conclude two things are different just because they look different. Often you have to do more work to find out whether they're the same or not. $\qquad$ $\endgroup$ – Michael Hardy Sep 14 '17 at 18:16

Substitute $e^{2y}$ by $x^3$ in the last line.


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