Answer to Question #139486 in Calculus for Harlene

Question

(xy)^x=e, solve for y'

Answer 1

Consider the equation $(xy)^x=e$

Take log of base $e$ both sides to obtain,

$ln(xy)^x=ln(e)$

$xln(xy)=1$ using $ln(a^n)=nln(a)$

Now differentiate $xln(xy)=1$ implicitly with respect to $x$ as,

$x\frac{d}{dx}ln(xy)+ln(xy)\frac{d}{dx}(x)=\frac{d}{dx}(1)$ using product rule

$x(\frac{y+x\frac{dy}{dx}}{xy})+ln(xy)=0$

Replace $\frac{dy}{dx}=y'$ and separate for $y'$ as,

$\frac{y+xy'}{y}+ln(xy)=0$

$y+xy'+yln(xy)=0$

$xy'=-y-yln(xy)$

$xy'=-y(1+ln(xy))$

$y'=\frac{-y(1+ln(xy))}{x}$

$y'=-\frac{y(1+ln(xy))}{x}$

Therefore, the first derivative with respect to $x$ is $y'=-\frac{y(1+ln(xy))}{x}$