`g(r) = r^2ln(2r + 1)` Differentiate the function.

Textbook Question

Chapter 3, 3.6 - Problem 14 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to differentiate the given function, with respect to "r" variable, such that:

`g'(r) = (r^2*ln(2r+1))'`

Using the product rule, yields:

`g'(r) = (r^2)'*ln(2r+1) + r^2*(ln(2r+1))'`

You need to use the chain rule to differentiate the function `ln(2r+1)` , such that:

`(ln(2r+1))' = (1/(2r+1))*(2r+1)'`

`(ln(2r+1))' = (2/(2r+1))`

`g'(r) = 2r*ln(2r+1) + (2r^2)/(2r+1)`

Factoring out 2r, yields:

`g'(r) = 2r*(ln(2r+1) + r/(2r+1))`

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

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