How do you find the derivative of ln(x^2)?

Jun 21, 2016

$\setminus \frac{2}{x}$

Explanation:

Using the chain rule and letting $u \left(x\right) = {x}^{2}$ we have $\setminus \frac{d}{\mathrm{dx}} \setminus \ln \left(u \left(x\right)\right) = \setminus \frac{d}{\mathrm{du}} \setminus \ln \left(u \left(x\right)\right) . \setminus \frac{\mathrm{du}}{\mathrm{dx}} = \setminus \frac{1}{u \left(x\right)} 2 x$

$= \setminus \frac{2 x}{{x}^{2}} = \setminus \frac{2}{x}$

Jun 21, 2016

$\frac{2}{x}$

Explanation:

There are two methods:

Using the Chain Rule:

Since the derivative of $\ln \left(x\right)$ is $\frac{1}{x}$, we see that the derivative of a function inside the natural logarithm, such as $\ln \left(f \left(x\right)\right)$, is $\frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$.

So, for $\ln \left({x}^{2}\right)$, the derivative is $\frac{1}{x} ^ 2 \cdot 2 x$, since $2 x$ is the derivative of ${x}^{2}$.

Then, we see that $\frac{1}{x} ^ 2 \cdot 2 x$ simplifies to $\frac{2}{x}$.

Simplifying first:

Using the rule $\log \left({a}^{b}\right) = b \cdot \log \left(a\right)$, we see that $\ln \left({x}^{2}\right) = 2 \cdot \ln \left(x\right)$.

The derivative of $2 \ln \left(x\right)$ is just $2$ times the derivative of $\ln \left(x\right)$, which is $\frac{1}{x}$.

We see that $2 \cdot \frac{1}{x} = \frac{2}{x}$, the answer we obtained earlier.