Homework Help

What is derivative of function y = (sin(2x^2-ln(2x^2)))^2?

user profile pic

jigsfix | Student, Undergraduate | (Level 3) eNoter

Posted August 11, 2013 at 2:21 PM via web

dislike 1 like

What is derivative of function y = (sin(2x^2-ln(2x^2)))^2?

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 11, 2013 at 3:18 PM (Answer #1)

dislike 1 like

You need to use the chain rule to evaluate the derivative of the function, hence, you need to start to differentiate the power function, then the sine function and then the argument of the sine function, such that:

`y' = 2sin(2x^2 - ln(2x^2))*(sin(2x^2 - ln(2x^2)))'*(2x^2 - ln(2x^2))`

`y' = 2sin(2x^2 - ln(2x^2))*(cos(2x^2 - ln(2x^2))*(4x - ((2x^2)')/(2x^2))`

`y' = 2sin(2x^2 - ln(2x^2))*(cos(2x^2 - ln(2x^2))*(4x -4x/(2x^2))`

Factoring out `4x` within the last factor yields:

`y' = 8x*sin(2x^2 - ln(2x^2))*(cos(2x^2 - ln(2x^2))*(1 -1/(2x^2))`

You need to notice that the arguments of sine and cosine functions are equal, hence, you may use the double angle formula, such that:

`2 sin theta*cos theta = sin 2 theta`

Reasoning by analogy, yields:

`y' = 4x*sin(2(2x^2 - ln(2x^2)))*(1 -1/(2x^2))`

Hence, evaluating the derivative of the given composed function, using the chain rule, yields `y' = 4x*sin(2(2x^2 - ln(2x^2)))*(1 -1/(2x^2)).`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes