What is the derivative of $\sqrt{x^{2} - 1}$ $sqrt( x^2-1)$ ?

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What is the derivative of $\sqrt{x^{2} - 1}$ $sqrt( x^2-1)$ ?

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Answer 1 · 2015-12-27T17:09:18

$\frac{x}{\sqrt{x^{2} - 1}}$ $x/sqrt(x^2-1)$

Explanation:

This is equivalent to ${(x^{2} - 1)}^{\frac{1}{2}}$ $(x^2-1)^(1/2)$ .

According to the chain rule:

$\frac{d}{d x} (u^{\frac{1}{2}}) = \frac{1}{2} u^{- \frac{1}{2}} \cdot \frac{d u}{d x}$ $d/dx(u^(1/2))=1/2u^(-1/2) * (du)/dx$

Thus,

$\frac{d}{d x} ({(x^{2} - 1)}^{\frac{1}{2}}) = \frac{1}{2} {(x^{2} - 1)}^{- \frac{1}{2}} \frac{d}{d x} (x^{2} - 1)$ $d/dx((x^2-1)^(1/2))=1/2(x^2-1)^(-1/2)d/dx(x^2-1)$

$\Rightarrow \frac{1}{2 \sqrt{x^{2} - 1}} \cdot 2 x$ $=>1/(2sqrt(x^2-1)) * 2x$

$\Rightarrow \frac{x}{\sqrt{x^{2} - 1}}$ $=>x/sqrt(x^2-1)$

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What is the derivative of $\sqrt{x^{2} - 1}$ $sqrt( x^2-1)$ ?

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Explanation:

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