How can I differentiate $(ye^x)^{\frac{1}{x}}=y^2$?

Question

I have following relation to differentiate: $$(ye^x)^{\frac{1}{x}}=y^2.$$

However, I got a bit confused: I first simplified: $y^{\frac{1}{x}}e^1=y^2$ and then differentiated, but that doesn't seems to approach the right answer.

Answer 1

The hard way:

$$(y^{1/x})'e=\frac exy'y^{(1/x)-1}-\ln y\ y^{1/x}\frac e{x^2}=(y^2)'=2yy'.$$

then $$y'=\frac{\ln y\ y^{1/x}\frac e{x^2}}{\frac exy^{(1/x)-1}-2y}.$$

Using $y^{1/x}=y^2/e$, this can be simplified to

$$\frac{\ln y\ \frac{y^2}e\frac e{x^2}}{\frac ex\frac{y^2}ey^{-1}-2y}=\frac{\ln y\ \frac{y^2}{x^2}}{\frac yx-2y}=\frac{\ln y\ y}{x(1-2x)}=\frac{(2\ln y-1)y}{1-2x}.$$

Answer 2

If the exponents and chain rule are giving you trouble together, you could take the logarithm of both sides first to eliminate the former--- $$\frac{\log y}{x} + 1 = 2 \log y$$ ---and then optionally rearrange: $$x = (2 x - 1) \log y.$$

Differentiating gives $$1 = 2 \log y + (2 x - 1)\frac{y'}{y}$$ and rearranging gives an expression for $y'$.

Answer 3

you are going to differentiate implicitly, you may as well take the $\ln$ of $y^{1/x}e = y^2$ and get $\dfrac{ln y}{x} + 1 = 2 \ln y $ which can be written as $$(2x -1)\ln y = x $$ which difference gives you $$(2x-1) \dfrac{dy}{y} + 2\ln y dx = dx$$ therefore $$\dfrac{dy}{dx} = \dfrac{y(2 \ln y - 1)}{1 - 2x} $$