`g(x) = (x^2)((2/x) - 1/(x+1))` Find the derivative of the algebraic function.

Textbook Question

Chapter 2, 2.3 - Problem 34 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to find derivative of the function using the product rule:

`f'(x)= ((x^2)'*(2/x - 1/(x+1)) + (x^2*(2/x - 1/(x+1))'`

`f'(x)= 2x*(2/x - 1/(x+1)) + (x^2)*(-2/(x^2) + 1/((x+1)^2))`

`f'(x)= 4 - (2x)/(x+1) - 2 + (x^2)/((x+1)^2)`

`f'(x)=2 - (2x)/(x+1)+ (x^2)/((x+1)^2)`

Factoring out `x/(x+1)` yields:

`f'(x)=2 -x/(x+1)(-2 + x/(x+1))`

`f'(x)=2 -x/(x+1)((-2x - 2+ x)/(x+1))`

`f'(x)=2+x(x+2)/((x+1)^2)`

Hence, evaluating the derivative of the function, yields `f'(x)=2+x(x+2)/((x+1)^2).`

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question