what is the derivative of $\ln(2x^2)$
Sophia Terry 
In finding the derivative of $\ln(2x^2),\;$ I have applied the chain rule and obtained $2x / x^2.\;$ Is this correct?
If not could some please explain how to do it?
$\endgroup$ 24 Answers
$\begingroup$If you use rules of logarithms, you don't even have to appeal to the chain rule.
Notice that $\ln\left(2x^{2}\right)=\ln\left(2\right)+2\ln\left(x\right)$, so the derivative is $\frac{2}{x}$.
$\endgroup$ $\begingroup$Another possible way is: if $x\neq 0$, then by setting $y=\ln(2x^2)$ we have $\exp(y)=2x^2$ so $y'e^y= (e^y)'=4x$ so $$y'=4xe^{-y}=4x\times\left(\frac{1}{2x^2}\right)=\frac{2}{x}$$
$\endgroup$ 3 $\begingroup$Using the chain rule, if you have a function written as $h(x) = f(g(x))$, its derivative is $f'(g(x))g'(x)$. For $h(x)=\ln(2x^2)$, you can take $f$ and $g$ to be $$ f(x) = \ln(x)\\ g(x) = 2x^2$$
Deriving them gives $f'(x) = 1/x$ and $g'(x) = 4x$. Insert this into the chain rule and you get $$ h'(x) = \frac{1}{2x^2}\cdot 4x = \frac{2}{x} $$
$\endgroup$ $\begingroup$You did just fine, and you correctly applied the chain rule. You correctly found the derivative. All you may want to do now is to simplify $$\require{cancel}\Big(\ln(2x^2)\Big)' = \frac {2x}{x^2} = \frac {2\cancel{x}}{x\cdot \cancel{x}} = \dfrac 2{x}$$
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