Find $y^{\prime }$ if $y=\ln (x^2\cos x)$

Step 1 ：Find the derivative of $y$ with respect to $x$ if $y=\ln (x^2\cos x)$

Step 2 ：Use the chain rule and the product rule for differentiation

Step 3 ：The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function

Step 4 ：The product rule states that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function

Step 5 ：Apply the chain rule and product rule to find $y^{\prime }$

Step 6 ：$y^{\prime }=\frac{d}{dx}\ln (x^2\cos x)$

Step 7 ：$y^{\prime }=\frac{1}{x^2\cos x}\cdot \frac{d}{dx}(x^2\cos x)$

Step 8 ：$y^{\prime }=\frac{1}{x^2\cos x}\cdot (x^2\cdot \frac{d}{dx}\cos x+\cos x\cdot \frac{d}{dx}{x}^2)$

Step 9 ：$y^{\prime }=\frac{1}{x^2\cos x}\cdot (x^2\cdot (-\sin x)+\cos x\cdot 2x)$

Step 10 ：$y^{\prime }=\frac{-x^2\sin x+2x\cos x}{x^2\cos x}$

Step 11 ：Simplify the final answer

Step 12 ：$\boxed{{\displaystyle \frac{-x^2\sin x+2x\cos x}{x^2\cos x}}}$