How do you differentiate ln x^(1/3)?

Jul 29, 2016

$\frac{1}{3 x}$

Explanation:

There are two methods: one that simplifies the function first, and the other which doesn't.

Without simplifying the function:

$y = \ln \left({x}^{\frac{1}{3}}\right)$

We will need to use the chain rule. Since:

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

The chain rule tells us that:

$\frac{d}{\mathrm{dx}} \ln \left(u\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Thus:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x} ^ \left(\frac{1}{3}\right) \cdot \frac{d}{\mathrm{dx}} {x}^{\frac{1}{3}}$

Use the power rule to find $\frac{d}{\mathrm{dx}} {x}^{\frac{1}{3}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x} ^ \left(\frac{1}{3}\right) \cdot \frac{1}{3} {x}^{- \frac{2}{3}}$

Simplify:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \cdot \frac{1}{x} ^ \left(\frac{1}{3}\right) \cdot \frac{1}{x} ^ \left(\frac{2}{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \cdot \frac{1}{x} ^ \left(\frac{3}{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3 x}$

Alternatively, simplify first:

You may remember a rule of logarithms that states that:

$\log \left({a}^{b}\right) = b \cdot \log \left(a\right)$

Thus:

$\ln \left({x}^{\frac{1}{3}}\right) = \frac{1}{3} \cdot \ln \left(x\right)$

So:

$y = \frac{1}{3} \cdot \ln \left(x\right)$

Now when differentiating, we won't have to use the chain rule, and the $\frac{1}{3}$ is simply brought out of the differentiation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \cdot \frac{d}{\mathrm{dx}} \ln \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \cdot \frac{1}{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3 x}$

Generalization:

This method can be applied to differentiating any function in the form:

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

Since $y = a \cdot \ln \left(x\right)$, we know that $\frac{\mathrm{dy}}{\mathrm{dx}} = a \cdot \frac{1}{x} = \frac{a}{x}$.

This can also be proven using the power rule, but above way is simpler.