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: $y'=y\left[\cos x\ln x+\frac{\sin x}x\right]=x^{\sin x} \left[\cos x\ln x+\frac{\sin x}x\right]$
 A: The derivative of $\ln y$ with respect to $y$ is certainly $1/y$, but in this case the derivative needs to be taken with respect to x, so by the chain rule, the derivative of $\ln y$ with respect to $x$ is $y'/y$. 
A: $$\left(f(g(x))\right)'=f'(g(x))g'(x).$$
Thus, $\left(\ln{y}\right)'=\frac{1}{y}\cdot{y'}$.
But I like the following way.
$$(x^{\sin{x}})'=\left(e^{\sin{x}\ln{x}}\right)'=x^{\sin{x}}\left(\cos{x}\ln{x}+\frac{\sin{x}}{x}\right).$$
A: $$
\frac {d\ln y}{dy} = \frac 1 y \qquad \text{ but } \qquad  \underbrace{ \frac {d\ln y}{dx} = \frac{d\ln y}{dy} \cdot  \frac{dy}{dx}}_{\large\text{This is the chain rule.}} = \frac 1 y \cdot y'.
$$
A: [Converted from comment]
What you are missing is saying "the derivative of $\ln y$ is $1/y$". This is incomplete. What is true is that "the derivative of $\ln y$ with respect to $y$ is $1/y$". But you are not differentiating with respect to $y$ -- you are differentiating with respect to $x$. The correct statement is "the derivative of $\ln y$ with respect to $x$ is $y'/y$". It's the chain rule.
A: $$y=x^{\sin x}\\\ln(y)=\ln(x^{\sin x}) \to \ln y = \sin x \ln x\\(\ln y = \sin x \ln x \to )'\\ \to 
\frac{y'}{y}=\cos x \cdot \ln x+\sin x.\cdot \frac 1x $$now to obtain y' 
you must multiply by y 
$$y\times(\frac{y'}{y}=\cos x \cdot \ln x+\sin x.\cdot \frac 1x ) \to \\
y'=y(\cos x \cdot \ln x+\sin x.\cdot \frac 1x )\\y'=x^{\sin x}\cdot(\cos x \cdot \ln x+\sin x.\cdot \frac 1x )$$
