Question

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

Answer 1

refer to answers below =)

Explanation:

The derivative of a `y= ln(f(x))` function is

`dy/dx= color(red)(f'(x))/color(blue)f(x)`

The equation is given as `y = lnsqrt(8x-4)`

Which can be changed to `y=ln(8x-4)^(1/2)`

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

`y=(1/2)ln(8x-4)`

Differentiating it,

`dy/dx= (1/2) color(blue)8/color(red)(8x-4)` Answer

Note that the numerator is `color(red)(f'(x))`.

Therefore, differentiating `color(blue)(f(x))=8x-4` results in `color(red)(f'(x))=8`.

Given `dy/dx= color(red)(f'(x))/color(blue)f(x)`,

We will get `dy/dx= 8/(8x-4). 1/2`

Answer 2

`dy/dx=4/(8x-4)`

Explanation:

`"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=lnsqrt(8x-4)=ln(8x-4)^(1/2)`

`dy/dx=1/((8x-4)^(1/2))xxd/dx(8x-4)^(1/2)`

`color(white)(dy/dx)=1/((8x-4)^(1/2))xx1/2(8x-4)^(-1/2).d/dx(8x-4)`

`color(white)(dy/dx)=1/((8x-4)^(1/2))xx4(8x-4)^(-1/2)`

`color(white)(dy/dx)=4/(8x-4)`