Differentiation of logarithmic functions using the chain rule What's the derivative of $x^2(\ln(x^2))$?
I'm having a really hard time with logarithmic differentiation. Can someone help rationalize it for me?
 A: In this case, we have an application of the product rule.
We start with
$$y = x^{2}(\ln(x^{2}))$$
To make this easier, let's rewrite this using the rules of logs as:
$$y = 2x^{2}(\ln(x))$$
Differentiating and applying the product rule, we find:
$$y' = (2x^{2})'(\ln(x)) + (2x^{2})(\ln(x))'$$
Since $$\frac{d}{dx} \ln(f(x)) = \frac{1}{f(x)}f'(x)$$
We have:
$$y' = 4x\ln(x) + 2x^{2}(\frac{1}{x})$$
So
$$y' = 4x\ln(x) + 2x = 2x(2\ln(x) + 1)$$
Edit: in the case that you wanted to do this via logarithmic differentiation (and you were not confused by the differentiation of logarithmic functions), here's how you would proceed:
We go back to 
$$y = 2x^{2}(\ln(x))$$
Then taking the natural log of both sides and using rules of logs:
$$\ln(y) = \ln(2x^{2}\ln(x)) = \ln(2x^{2}) + \ln(\ln(x)) = 2\ln(2x) + \ln(\ln(x))$$
Again, using rules for differentiation of logs and the chain rule we find:
$$\frac{1}{y}y' = 2\cdot(\frac{1}{2x}\cdot 2) + (\frac{1}{\ln x}*\frac{1}{x})$$
Which gives us
$$\frac{y'}{y} = \frac{2}{x} + \frac{1}{x\ln x}$$
So just as we found before:
$$y' = (\frac{2}{x} + \frac{1}{x\ln x})(2x^{2}\ln(x)) = 4x\ln(x) + 2x = 2x(2\ln(x) + 1)$$
A: Hints:


*

*$\bigl( f \cdot g \bigr)' = f \cdot g' + g \cdot f'$.


Using this we see that if we see that 
\begin{align*}y &= x^{2} \cdot \log(x^{2})\\
\implies y' &= x^{2} \cdot \frac{d}{dx}(\log(x^{2})) + \log(x^{2}) \cdot \frac{d}{dx} (2x) \\ &= x^{2} \cdot \frac{d}{dx} (2\cdot \log(x)) + \log(x^{2}) \times 2 \\
&=2 \cdot x^{2} \cdot \frac{1}{x} \cdot \log(x) + 4 \cdot \log(x)
\\ &= 2x \cdot \log(x) + 4 \cdot \log(x)
\end{align*}
