Logarithmic Differentiation equation. I have to differentiate this via log. I am still learning, so please be patient, I will try to explain everything I did. Please tell me if it is correct. y=((x+3)^4(2x^2+5x)^3)/(sqrt(4x-3)) ln(y) = ln((x+3)^4(2x^2+5x)^3)) - ln(sqrt(4x-3))

Kyler Oconnor

Kyler Oconnor

Answered question

2022-11-06

Logarithmic Differentiation equation, Help!
So, I have to differentiate this via log. I am still learning, so please be patient, I will try to explain everything I did. Please tell me if it is correct.
y = ( x + 3 ) 4 ( 2 x 2 + 5 x ) 3 4 x 3
ln ( y ) = ln ( ( x + 3 ) 4 ( 2 x 2 + 5 x ) 3 ) ) ln ( 4 x 3 )
ln ( y ) = 4 ln ( x + 3 ) 3 ln ( 2 x 2 + 5 x ) 1 2 ln ( 4 x 3 )
aaaaaaaaand don't know what to do next, any help in the process or next step?

Answer & Explanation

reinmelk3iu

reinmelk3iu

Beginner2022-11-07Added 21 answers

taking th derivative of the last equation you have, we get
1 y d y d x = 4 x + 3 + 3 ( 4 x + 5 ) 2 x 2 + 5 x 2 4 x 3
sbrigynt7b

sbrigynt7b

Beginner2022-11-08Added 2 answers

You've left out a plus sign in the last line. It should read
ln ( y ) = 4 ln ( x + 3 ) + 3 ln ( 2 x 2 + 5 x ) 1 2 ln ( 4 x 3 ) .
Then you can still factor the second term to get
ln ( y ) = 4 ln ( x + 3 ) + 3 ln ( x ( 2 x + 5 ) ) 1 2 ln ( 4 x 3 ) = 4 ln ( x + 3 ) + 3 ln ( x ) + 3 ln ( 2 x + 5 ) 1 2 ln ( 4 x 3 ) . <>rbFrom that point, you can go ahead and differentiate to obtain
y y = 4 x + 3 + 3 x + 6 2 x + 5 2 4 x 3 .
Then multiply the y across to get
y = ( 4 x + 3 + 3 x + 6 2 x + 5 2 4 x 3 ) y .
Finally, you would like to express y′ as a function of x, rather than both x and y, so substitute in your original equation
y = ( x + 3 ) 4 ( 2 x 2 + 5 x ) 3 4 x 3
on the right to obtain
y = ( 4 x + 3 + 3 x + 6 2 x + 5 2 4 x 3 ) ( ( x + 3 ) 4 ( 2 x 2 + 5 x ) 3 4 x 3 ) ,
and then simplify to taste.

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?