Question

(3) Find the equation of the tangent line to $y = \frac{x^2 - 1}{x^2 + 1}$ at $x = 0$.

Name: (3) Find the equation of the tangent line to y = (x^2 - 1)/(x^2 + 1) at x = 0.
Uploaded: 2023-12-30T10:42:39-08:00
Duration: 4 min 35 s
Channel: Adi S
Description: (3) Find the equation of the tangent line to y = (x^2 - 1)/(x^2 + 1) at x = 0.

          (3) Find the equation of the tangent line to $y = \frac{x^2 - 1}{x^2 + 1}$ at $x = 0$.

Added by Cindy M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

12/30/2023

Step 1

The derivative of y with respect to x is given by: y' = [(2x)(x^2 + 1) - (x^2 - 1)(2x)] / (x^2 + 1)^2 y' = (2x^3 + 2x - 2x^3 + 2x) / (x^2 + 1)^2 y' = 4x / (x^2 + 1)^2 Show more…

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