# How do you use logarithmic differentiation to find the derivative of the function?

Mar 28, 2015

Logarithmic differentiation when the logarithm of an expression can be written in a form for which we can find the derivative more simply. Or, for which we can find the derivative at all.
Here's one of each:

Example 1
For $y = {\left(3 x + 5\right)}^{5} / {\left(4 x - 9\right)}^{7}$, we could find $\frac{\mathrm{dy}}{\mathrm{dx}}$ by using the quotient rule, the power rule and the chain rule and then simplify algebraically.
OR
$\ln y = \ln \left({\left(3 x + 5\right)}^{5} / {\left(4 x - 9\right)}^{7}\right) = 5 \ln \left(3 x + 5\right) - 7 \ln \left(4 x - 9\right)$

Differentiating implicitly gives us:

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 5 \left(\frac{1}{3 x + 5} \cdot 3\right) - 7 \left(\frac{1}{4 x - 9} \cdot 4\right)$.

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{15}{3 x + 5} - \frac{28}{4 x - 9}$

Remembering that $y = {\left(3 x + 5\right)}^{5} / {\left(4 x - 9\right)}^{7}$, we can solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{15}{3 x + 5} - \frac{28}{4 x - 9}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(3 x + 5\right)}^{5} / {\left(4 x - 9\right)}^{7} \left(\frac{15}{3 x + 5} - \frac{28}{4 x - 9}\right)$

Example 2
$y = \sqrt[x]{x} = {x}^{\frac{1}{x}}$.

This is neither an exponential, nor a power function. we need some other way to find $\frac{\mathrm{dy}}{\mathrm{dx}}$. logarithmic differentiation will work.

$\ln y = \frac{1}{x} \ln x$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x} ^ 2 \ln x + \frac{1}{x} \cdot \frac{1}{x} = \frac{1 - \ln x}{x} ^ 2$

Solving gives:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt[x]{x} \left(1 - \ln x\right)}{x} ^ 2$