Differentiate:$ x^{2y} = \ln y$

Question

I am having trouble with the following equation defined implicitly with respect to $x$.I figured I could use the quotient rule or maybe play around with the logs, however I always need to differentiate $x^{2y}$ which I have no idea how to do, any help will be appreciated.

Differentiate $x^{2y} = \ln y$, with respect to $x$.

Answer 1

Assuming $y(x)= f(x)$. $$\begin{align}e^{\ln (x^{2y})} = \ln y &\Rightarrow e^{2y \ln x} = \ln y \Rightarrow e^{2y \ln x}\Big(\frac{2y}{x} + 2y'\ln x\Big) = \frac{y'}{y} \\&\Rightarrow y' - 2yy'\ e^{2y \ln x} \ln x= \frac{2y^2}{x}e^{2y \ln x} \Rightarrow y' = \frac{2y^2\ \ e^{2y \ln x}}{x(1 - 2y\ e^{2y \ln x} \ln x)}\end{align}$$

Answer 2

Inplicit differentiation:

$$x^{2y}=\log y\implies 2yx^{2y-1}\,dx+2x^{2y}\log x\,dy=\frac1ydy\implies\left(\frac1y-2x^{2y}\log x\right)dy=2yx^{2y-1}\,dx\implies$$

$$\frac{dy}{dx}=\frac{2yx^{2y-1}}{\frac1y-2x^{2y}\log x}=\frac{2x^{2y-1}y^2}{1-2x^{2y}y\log x}$$

The answer was edited since the OP was.

Answer 3

we have $x^{2y}=\ln(y)$ rewriting this as $e^{2y\ln(x)}=\ln(y)$ now differentiaitng with respect to $x$: $e^{2y\ln(x)}(2y'\ln(x)+\frac{2y}{x})=\frac{1}{y}{y'}$