help with logs (1 Viewer)

[simDS]

hey i cant do this question could someone please solve.

f(x)= loge (2x+1) find f`(1)

thanx

 

[Revacious]

f'x = 2/2x+1

f'1 = 2/3

 

[simDS]

wait i dont het how to differentiate it
how did you do it?

 

[kurt.physics]

hey i cant do this question could someone please solve.

f(x)= loge (2x+1) find f`(1)

thanx

 

[samthebear]

when you're differenciating a log the rule goes:

dy/dx loge f(x) = f'(x) / f(x)

in words:

the differencial of a log function is the derivative of the function divided by the original function.

 

[Revacious]

i never remembered learning it that way =\

i learned it as chain rule, which i suppose is the same thing


y=ln(2x+1) let u = 2x+1
dy/dx=dy/du times du/dx

dy/du is 1/u and du/dx is 2

therefore dy/dx = 2/u

which is 2/2x+1

 

[kurt.physics]

i never remembered learning it that way =\

i learned it as chain rule, which i suppose is the same thing


y=ln(2x+1) let u = 2x+1
dy/dx=dy/du times du/dx

dy/du is 1/u and du/dx is 2

therefore dy/dx = 2/u

which is 2/2x+1

The formula we mentioned was derived from generalizing that particular function then differentiating with the chain rule.