Question

Use logarithmic differentiation to find dy/dx

Answer 1

Answer:

Hello,

Step-by-step explanation:

Let's say:

$y=x^{cotg(x)}\\\\ln(y)=cotg(x)*ln(x)\\\\\\\dfrac{y'}{y}=\dfrac{-1}{sin^2(x)} *ln(x)+\dfrac{cotg(x)}{x} \\\\\\y'=-\dfrac{x^{cotg(x)}*ln(x)}{sin^2(x)} +cotg(x)*x^{cotg(x)-1}$