# How do you differentiate y = lnx^2?

Mar 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x}$

#### Explanation:

Applying the chain rule, along with the derivatives $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$ and $\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$, we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \ln \left({x}^{2}\right)$

$= \frac{1}{x} ^ 2 \left(\frac{d}{\mathrm{dx}} {x}^{2}\right)$

$= \frac{1}{x} ^ 2 \left(2 x\right)$

$= \frac{2}{x}$

Mar 4, 2016

$\frac{2}{x}$

#### Explanation:

Alternatively, we can simplify $\ln \left({x}^{2}\right) = 2 \ln \left(x\right)$ from the outset, using the rule that $\log \left({a}^{b}\right) = b \log \left(a\right)$.

Since $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$, we see the constant can be brought from the differentiation in $\frac{d}{\mathrm{dx}} 2 \ln \left(x\right) = 2 \frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{2}{x}$.

Mar 4, 2016

$\frac{2}{x}$

#### Explanation:

Just to show the versatility of calculus, we can solve this problem through implicit differentiation.

Raise both side to the power of $e$.

$y = \ln \left({x}^{2}\right)$

${e}^{y} = {e}^{\ln} \left({x}^{2}\right)$

${e}^{y} = {x}^{2}$

Differentiate both sides with respect to $x$. The left side will require the chain rule.

${e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{e} ^ y$

Recall that ${e}^{y} = {x}^{2}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{x} ^ 2 = \frac{2}{x}$