Problem

Find an equation of the tangent line to the given…

01:52

Problem 30 Medium Difficulty

Find $ f'(x) $ and $ f"(x). $

$ f(x) = \frac {x}{x^2 - 1} $

Answer

$f^{\prime}(x)=\frac{-x^{2}-1}{\left(x^{2}-1\right)^{2}} \quad f^{\prime \prime}(x)=\frac{2 x\left(x^{2}+3\right)}{\left(x^{2}-1\right)^{3}}$

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Video Transcript

it's clear. So when you read here, so we have f of X is equal to X over X square minus one. We're gonna use the quotient role. We'll get X square minus one D acts d over DX minus x times D over D X X square minus one well over X square minus one square just becomes equal to X square minus one minus two X square over X square minus one square, which is equal to negative X square minus one over X square minus one square. We're gonna find the second derivative by using the Kocian rule, which is X square minus one square D over DX your negative X square minus one minus negative X square minus one de over de x X square minus one square All over X square minus one square. You square that again, this becomes equal to that's square minus one times negative. Two acts less negative. Two X square minus two turns negative. Two acts well over X square minus one. Cute. You could simplify this to two x times X square plus three all over X square minus one. Cute

Clarissa N.

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Section 2

The Product and Quotient Rules

Problem

Find an equation of the tangent line to the given…

Problem 30 Medium Difficulty

Find $ f'(x) $ and $ f"(x). $

$ f(x) = \frac {x}{x^2 - 1} $

Answer

$f^{\prime}(x)=\frac{-x^{2}-1}{\left(x^{2}-1\right)^{2}} \quad f^{\prime \prime}(x)=\frac{2 x\left(x^{2}+3\right)}{\left(x^{2}-1\right)^{3}}$

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Calculus: Early Transcendentals

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