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# Find $f'(x)$ and $f"(x).$$f(x) = \frac {x}{x^2 - 1}$

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

Derivatives

Differentiation

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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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

#### Topics

Derivatives

Differentiation

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp