implicit function with differential:y(x)=x^y(x) help

**jmanbb** · August 10th 2012, 03:41 AM

Hello,in a textbook exercise,the solution to the function y(x)=x^y(x) is given with differential.We set Xo=1,therefore y(xo)=y(1)=1.Then dy=y'(xo)dx=>dy=y'(1)*(1.1-1).Then it uses implicit factoring, y'(1)=y^2(1)/1/1-y(1)ln1=1.How do we get from y(1)=x^y(x to y'(1)=y^2(1)/1/1-y(1)ln1.I'm obviously missing something in terms of factoring,can someone help me?PS:The exercise's main goal is to find y(1.1)

**emakarov** · August 10th 2012, 03:57 AM

If $y(x) = x^{y(x)}=e^{y(x)\ln(x)}$ , then $y'(x)=x^{y(x)}(y'(x)\ln(x) + y(x)/x) = y^2(x)/x+y(x)y'(x)\ln(x)$ . Solve this equation for y'(x) and make sure you use parentheses where necessary.

**Prove It** · August 10th 2012, 04:07 AM

$\displaystyle \begin{align*} y &= x^y \\ \ln{y} &= \ln{\left(x^y\right)} \\ \ln{y} &= y\ln{x} \\ \frac{\ln{y}}{y} &= \ln{x} \\ \frac{d}{dx}\left(\frac{\ln{y}}{y}\right) &= \frac{d}{dx}\left(\ln{x}\right) \\ \frac{d}{dy}\left(\frac{\ln{y}}{y}\right)\frac{dy} {dx} &= \frac{1}{x} \\ \left[\frac{y\left(\frac{1}{y}\right) - 1\ln{y}}{y^2}\right]\frac{dy}{dx} &= \frac{1}{x} \\ \left(\frac{1 - \ln{y}}{y^2}\right)\frac{dy}{dx} &= \frac{1}{x} \\ \frac{dy}{dx} &= \frac{y^2}{x\left(1 - \ln{y}\right)} \end{align*}$

**jmanbb** · August 10th 2012, 04:15 AM

thank you both for your answers ,they're really helpful!

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