Question

For the following exercises, use logarithmic differentiation to find dy/dx.

Answer 1

If $y=x^{\log_2(x)}$, then taking the logarithm of both sides gives

$\ln(y) = \ln\left(x^{\log_2(x)}\right) = \log_2(x) \ln(x)$

Differentiate both sides with respect to x :

$\dfrac{d\ln(y)}{dx} = \dfrac{d\log_2(x)}{dx}\ln(x) + \log_2(x)\dfrac{d\ln(x)}{dx}$

$\dfrac1y \dfrac{dy}{dx} = \dfrac{d\log_2(x)}{dx}\ln(x) + \dfrac{\log_2(x)}{x}$

Now if $z=\log_2(x)$, then $2^z=x$. Rewrite

$2^z = e^{\ln(2^z)} = e^{\ln(2)z}$

Then by the chain rule,

$\dfrac{d2^z}{dx} = \dfrac{dx}{dx}$

$\dfrac{de^{\ln(2)z}}{dx} = 1$

$e^{\ln(2)z} \ln(2) \dfrac{dz}{dx}= 1$

$\dfrac{dz}{dx} = \dfrac{1}{e^{\ln(2)z}\ln(2)}$

$\dfrac{dz}{dx} = \dfrac{1}{2^z \ln(2)}$

$\dfrac{d\log_2(x)}{dx} = \dfrac{1}{\ln(2)x}$

So we have

$\dfrac1y \dfrac{dy}{dx} = \dfrac{\ln(x)}{\ln(2)x}+ \dfrac{\log_2(x)}{x}$

$\dfrac1y \dfrac{dy}{dx} = \dfrac{\log_2(x)}{x}+ \dfrac{\log_2(x)}{x}$

$\dfrac1y \dfrac{dy}{dx} = \dfrac{2\log_2(x)}{x}$

$\dfrac1y \dfrac{dy}{dx} = \dfrac{\log_2(x^2)}{x}$

$\dfrac{dy}{dx} = \dfrac{y\log_2(x^2)}{x}$

Replace y :

$\dfrac{dy}{dx} = \dfrac{x^{\log_2(x)}\log_2(x^2)}{x}$

$\dfrac{dy}{dx} = x^{\log_2(x)-1}\log_2(x^2)$