# How do you find the derivative of y=lnsqrt(8x-4)?

May 18, 2017

#### Explanation:

The derivative of a $y = \ln \left(f \left(x\right)\right)$ function is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\textcolor{red}{f ' \left(x\right)}}{\textcolor{b l u e}{f}} \left(x\right)$

The equation is given as $y = \ln \sqrt{8 x - 4}$

Which can be changed to $y = \ln {\left(8 x - 4\right)}^{\frac{1}{2}}$

The properties of a natural log function allows for its exponent to be "brought" down as such:

$y = \left(\frac{1}{2}\right) \ln \left(8 x - 4\right)$

Differentiating it,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{2}\right) \frac{\textcolor{b l u e}{8}}{\textcolor{red}{8 x - 4}}$ Answer

Note that the numerator is $\textcolor{red}{f ' \left(x\right)}$.

Therefore, differentiating $\textcolor{b l u e}{f \left(x\right)} = 8 x - 4$ results in $\textcolor{red}{f ' \left(x\right)} = 8$.

Given $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\textcolor{red}{f ' \left(x\right)}}{\textcolor{b l u e}{f}} \left(x\right)$,

We will get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8}{8 x - 4} . \frac{1}{2}$

May 18, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{8 x - 4}$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

• d/dx(ln(f(x)))=(f'(x))/(f(x))

• d/dx(f(g(x))=f'(g(x))xxg'(x)larr" for "sqrt(8x-4)

$y = \ln \sqrt{8 x - 4} = \ln {\left(8 x - 4\right)}^{\frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\left(8 x - 4\right)}^{\frac{1}{2}}} \times \frac{d}{\mathrm{dx}} {\left(8 x - 4\right)}^{\frac{1}{2}}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{1}{{\left(8 x - 4\right)}^{\frac{1}{2}}} \times \frac{1}{2} {\left(8 x - 4\right)}^{- \frac{1}{2}} . \frac{d}{\mathrm{dx}} \left(8 x - 4\right)$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{1}{{\left(8 x - 4\right)}^{\frac{1}{2}}} \times 4 {\left(8 x - 4\right)}^{- \frac{1}{2}}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{4}{8 x - 4}$