`y = x^2 ln(2x)` Find y' and y''.

Textbook Question

Chapter 3, 3.6 - Problem 23 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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kalau's profile pic

kalau | (Level 2) Adjunct Educator

Posted on

  Find y' and y''.

For the first derivative, use the product rule AB'+BA'.  

Be warned that since we have a ln(2x) term, we will need to use chain rule, which is the derivative of the inner term 2x.

`y' = x^2* (1/(2x)) * 2 + ln(2x) *2x`

Simplify the above equation.

The first derivative is:  `y'=x + 2xln(2x)`

To find the second derivative, we will need to take the derivative of each term.  The derivative of first term is simply 1.  The second term will require product rule and chain rule like what we have done for the first derivative.

`y'=x+[(2x)(ln(2x))]`

Apply the derivative.

`y'' = 1 + [(2x)(1/ (2x))*2 + ln(2x) (2)]`

Simplify.

`y''= 1+2+2ln(2x)`

The second derivative is:  `y''= 2ln(2x)+3`

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to differentiate the function with respect to x, using product rule and chain rule, such that:

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

`y' = (x^2)'*ln(2x) + x^2*(ln(2x))'`

`y' =2x*ln(2x) + x^2*1/(2x)*(2)`

Factoring out `2x` , yields:

`y' =2x*(ln(2x) + x/(2x))`

`y' =2x*(ln(2x) + 1/2)`

Hence, evaluating the derivative of the given function, yields `y' =2x*(ln(2x) + 1/2).`

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