Question

Help please

Use logarithmic differentiation to find the derivative of y=(cosx)^tanx

Answer 1

123, ABC, 098, ZYX

Answer 2

Call me ;) ^

Answer 3

so, if I understand the terminology, this wants you to use properties of logs to make differentiation easier

Answer 4

so, what do you think your first step should be?
\[y=(cosx)^{tanx}\]

Answer 5

@sasha.o

Answer 6

\(\large\color{black}{ \displaystyle y=f(x)^{g(x)} }\)

To find the derivative (dy/dx), this is what you do:
(Abstract example, for some differentiable functions f and g)



\(\bf \color{red}{[1]}\) Hit both sides with a natural log, ln.

\(\large\color{black}{ \displaystyle \ln\left(y\right)=\ln\left(f(x)^{g(x)}\right) }\)



\(\bf \color{red}{[2]}\) Use \(\ln(A^B)=B\ln(A)\).

\(\large\color{black}{ \displaystyle \ln\left(y\right)={g(x)}\ln\left(f(x)\right) }\)



\(\bf \color{red}{[3]}\) Differentiate both sides, and remember that y will get a chain
rule of y' since y is really here to denote a function of x. And for the
left side you get use product rule \(dy/dx\)(f•g)=f\('\)g+f g\('\)

\(\large\color{black}{ \displaystyle \frac{y'}{y}=g'(x)\ln\left(f(x)\right)+g(x)\frac{f'(x)}{f(x)} }\)

\(\large\color{black}{ \displaystyle \frac{y'}{y}=g'(x)\ln\left(f(x)\right)+\frac{f'(x)g(x)}{f(x)} }\)



\(\bf \color{red}{[4]}\) Multiply both sides times y.

\(\large\color{black}{ \displaystyle y'=y\left[g'(x)\ln\left(f(x)\right)+\frac{f'(x)g(x)}{f(x)}\right] }\)



\(\bf \color{red}{[5]}\) You are given the function y in the beginning.

\(\large\color{black}{ \displaystyle y'=f(x)^{g(x)}\left[g'(x)\ln\left(f(x)\right)+\frac{f'(x)g(x)}{f(x)}\right] }\)