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How do you find the derivative of $g (x) = \sqrt{x^{2} - 1}$ $g(x) = sqrt(x^2-1)$ ?

1 Answer

Shwetank Mauria

Mar 21, 2016

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

Explanation:

We have to use here the concept of function of a function which uses the formula for chain rule.

As $g (x) = \sqrt{h (x)}$ $g(x)=sqrt(h(x)$ where $h (x) = x^{2} - 1$ $h(x)=x^2-1$ and according to chain rule

$\frac{d g}{d x} = \frac{d g}{d x} \times \frac{d h}{d x}$ $(dg)/(dx)=(dg)/dxxx(dh)/(dx)$

Hence $\frac{d g}{d x} = \frac{1}{2 \sqrt{x^{2} - 1}} \times 2 x$ $(dg)/(dx)=1/(2sqrt(x^2-1))xx2x$

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

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