# How do you find the derivative of y = lnx^2?

Jun 29, 2015

Either use the chain rule and $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$, or use properties of the logarithm to rewrite in simpler form.

#### Explanation:

$y = \ln {x}^{2}$

I assume that we are using correct notation and the function here is $f \left(x\right) = \ln \left({x}^{2}\right)$
(If we meant the square of the $\ln$ we would have to write ${\left(\ln x\right)}^{2}$ or, perhaps ${\ln}^{2} x$.)

Chain Rule Solution

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

Rewrite Solution

Use $\ln {a}^{r} = r \ln a$, to get:

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

Jul 1, 2015

Alternatively, if you have some free time, you can do some manipulation to this and get an idea of what it means to implicitly differentiate.

${e}^{y} = {x}^{2}$
Now we have $z = z \left(y \left(x\right)\right) = {e}^{y} = {x}^{2}$
where $z \left(y\right) = {e}^{y}$ and $y \left(x\right) = {x}^{2}$.

And $\frac{\mathrm{dz}}{\mathrm{dx}} = \left(\frac{\mathrm{dz}}{\mathrm{dy}}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Thus:
$\frac{\mathrm{dz}}{\mathrm{dx}} = \left(\frac{d \left[{e}^{y}\right]}{\mathrm{dy}}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

So now we just get:

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