Question

If x^(y)=5^(x-y) then dy/dx=​

Answer 1

First, rewrite

$x^y=e^{\ln x^y}=e^{y\ln x}$

$5^{x-y}=e^{\ln5^{x-y}} = e^{\ln(5)(x-y)}$

Now, differentiate both sides using the chain rule:

$\dfrac{\mathrm d\left(e^{y\ln x}\right)}{\mathrm dx}=\dfrac{\mathrm d\left(e^{\ln(5)(x-y)}\right)}{\mathrm dx}$

$e^{y\ln x}\dfrac{\mathrm d(y\ln x)}{\mathrm dx}=e^{\ln(5)(x-y)}\dfrac{\mathrm d(\ln(5)(x-y))}{\mathrm dx}$

$x^y\left(\dfrac{\mathrm dy}{\mathrm dx}\ln x+y\dfrac{\mathrm d(\ln x)}{\mathrm dx}\right)=\ln(5)5^{x-y}\left(\dfrac{\mathrm d(x)}{\mathrm dx}-\dfrac{\mathrm dy}{\mathrm dx}\right)$

$x^y\left(\ln x\dfrac{\mathrm dy}{\mathrm dx}+\dfrac yx\right)=\ln(5)5^{x-y}\left(1-\dfrac{\mathrm dy}{\mathrm dx}\right)$

$\left(x^y\ln x+\ln(5)5^{x-y}\right)\dfrac{\mathrm dy}{\mathrm dx}=\ln(5)5^{x-y}-yx^{y-1}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\ln(5)5^{x-y}-yx^{y-1}}{x^y\ln x+\ln(5)5^{x-y}}$