Logarithmic differentiation trouble with bottom of fraction y=((2x+3)^9)/(sqrt x(x^2-x)^6)

Hunter Shah

Hunter Shah

Answered question

2022-10-22

Logarithmic differentiation trouble with bottom of fraction
y = ( 2 x + 3 ) 9 x ( x 2 x ) 6
I switched it to ln ( y ) = 9 ln ( 2 x + 3 ) 6 x 1 / 2 ln ( x 2 x ) and then used the log rules for derivatives I know and the product rule on the right side and wound up with
y = y [ 18 2 x + 3 3 ln ( x 2 x ) x 1 / 2 + 6 x 1 / 2 ( 2 x 1 ) x 2 x ]
The 18 2 x + 3 is the correct form but the rest is way off from the form my answer is supposed to be in. I'm not quite sure what else to do with it though...

Answer & Explanation

Cavalascamq

Cavalascamq

Beginner2022-10-23Added 21 answers

To solve for the derivative of y = ( 2 x + 3 ) 9 x ( x 2 x ) 6 , we'll be using logarithmic differentiation which, as you said, you were having trouble with. In taking the natural logarithm on both sides, we now have
ln ( y ) = 9 ln ( 2 x + 3 ) ln ( x 1 / 2 ( x 2 x ) 6 ) .
We know the product logarithm rule, ln ( a b ) = ln ( a ) + ln ( b ), so we'll use that on our second term in the RHS:
ln ( y ) = 9 ln ( 2 x + 3 ) ( ln ( x 1 / 2 ) + ln ( ( x 2 x ) 6 ) .
We'll simplify down some of our RHS terms now:
ln ( y ) = 9 ln ( 2 x + 3 ) 1 / 2 ln ( x ) 6 ln ( x 2 x ) .
6 ln ( x 2 x ) can be simplified to 6 ln ( x ( x 1 ) ) , or 6 ln ( x ) 6 ln ( x 1 ) . We can now obtain
ln ( y ) = 9 ln ( 2 x + 3 ) 1 / 2 ln ( x ) 6 ln ( x ) 6 ln ( x 1 ) .
And I'll simplify 1 / 2 ln ( x ) 6 ln ( x ) = 13 / 2 ln ( x )
ln ( y ) = 9 ln ( 2 x + 3 ) 13 / 2 ln ( x ) 6 ln ( x 1 ) .
Using logarithmic differentials, we can do this:
y = y ( 18 2 x + 3 13 / 2 x 6 x 1 )
Substitute for y, giving
y = ( 2 x + 3 ) 9 x ( x 2 x ) 6 ( 18 2 x + 3 13 / 2 x 6 x 1 ) .
I'll let you do the rest.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?