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

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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)).`