`g(x) = (2ra^(rx) + n)^p` Find the derivative of the function.

Textbook Question

Chapter 3, 3.4 - Problem 43 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsenviro's profile pic

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

`g'(x) = d/dx (2ra^(rx)+n)^p`

`= p(2ra^(rx)+n)^(p-1) * d/dx (2ra^(rx)+n)`

`= p(2ra^(rx)+n)^(p-1) * 2r*ln a*a^(rx)*r`

`= 2pr^2 a^(rx)*ln(a)*(2ra^(rx)+n)^(p-1)`

Hope this helps.

balajia's profile pic

balajia | College Teacher | (Level 1) eNoter

Posted on

`g(x)=(2ra^(rx)+n)^p`

`g'(x)=p(2ra^(rx)+n)^(p-1).(2r).ra^(rx)loga`

`=2pr^2loga.a^(rx)(2ra^(rx)+n)^(p-1)`

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