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

2 Answers
John C.
May 18, 2017

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#

Jim G.
May 18, 2017

#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)#

Related questions
See all questions in Chain Rule
Impact of this question
2506 views around the world
You can reuse this answer
Creative Commons License