Find the derivative of $y=x^{\sin x}$ Could someone please explain step 3 for the following: Why do they multiply $1/y$ with $y'$? I understand that the derivative of $\ln y$ is $1/y$, but I don't understand why it is multiplied with $y'$ in step 3. 
Find the derivative for    $y=x^{\sin x}$
Step 1: $\ln y=\ln x^{\sin x}$
Step 2: $\ln y=\sin x\ln x$
Step 3: $\frac{y'}y=\cos x\ln x+\frac{\sin x}x$
Step 4: $y'=y\left[\cos x\ln x+\frac{\sin x}x\right]=x^{\sin x} \left[\cos x\ln x+\frac{\sin x}x\right]$

$$\left(f(g(x))\right)'=f'(g(x))g'(x).$$
Thus, $\left(\ln{y}\right)'=\frac{1}{y}\cdot{y'}$.
But I like the following way.
$$(x^{\sin{x}})'=\left(e^{\sin{x}\ln{x}}\right)'=x^{\sin{x}}\left(\cos{x}\ln{x}+\frac{\sin{x}}{x}\right).$$
