Use logarithmic differentiation to find the derivative of y = (cos(x))^ln(x)

Question

Okay so this question is on my exam study guide for Calculus 1, and I'm having trouble getting to an answer. Here are the steps I have tried currently.

1. Given: $y = (cos(x))^ln(x)$
2. Logarithmic differentiation: $ln(y) = ln(cos(x)^ln(x))$
3. Simplify: $(1/y)*(dy/dx) = ln(x) * ln(cos(x))$
4. Simplify: $(1/y)*(dy/dx) = (1/x) * ln(cos(x))$
5. Product Rule: $(1/y)*(dy/dx) = (1/x) * ln(cos(x)) + (ln(cos(x)))' * ln(x)$
6. Chain Rule: $(1/y)*(dy/dx) = (1/x)*ln(cos(x)) + [1/cos(x) * -sin(x) * ln(x)]$
7. Answer: $(dy/dx) = y((1/x)*ln(cos(x)) + [1/cos(x) * -sin(x) * ln(x)])$

Can someone help as to see where I went wrong in my thinking here

Answer 1

we get $$\frac{y'}{y}=\frac{1}{x}\ln(\cos(x))+\ln(x)\cdot \frac{1}{\cos(x)}\cdot (-\sin(x))$$

Answer 2

$$\begin{align}y=(\cos x)^{\ln x}&\implies \ln y=\ln x\ln(\cos x)\quad\boxed{\text{take logs and use that}\, \ln(a^b)=b\ln a}\\&\implies \overbrace{\frac1y\cdot\frac{dy}{dx}}^{\text{chain rule}}=\overbrace{\frac1x\ln(\cos x)+\ln x\cdot\left(\underbrace{\frac1{\cos x}\cdot(-\sin x)}_{\text{chain rule}}\right)}^{\text{product rule}}\\&\implies \frac{dy}{dx}=(\cos x)^{\ln x}\left(\frac{\ln(\cos x)}x-\tan x\ln x\right)\quad\boxed{\text{simplify}}\end{align}$$