What is the derivative of $\frac{\sqrt{x - 1}}{\sqrt{x}}$ $sqrt(x - 1)/sqrtx$ ?

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What is the derivative of $\frac{\sqrt{x - 1}}{\sqrt{x}}$ $sqrt(x - 1)/sqrtx$ ?

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Answer 1 · 2016-01-27T05:05:03

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

Explanation:

To find the derivative of $\frac{\sqrt{x - 1}}{\sqrt{x}}$ $sqrt(x-1)/sqrt(x)$ we can try the following
$\frac{\sqrt{x - 1}}{\sqrt{x}} = \sqrt{\frac{x - 1}{x}}$ $sqrt(x-1)/sqrt(x) = sqrt((x-1)/x)$
on simplifying further

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

Let us use chain rule.

Let $y = \sqrt{u}$ $y=sqrt(u)$ and $u = 1 - \frac{1}{x}$ $u=1-1/x$

By chain rule

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $dy/dx = dy/(du) xx (du)/dx$

$y = \sqrt{u}$ $y=sqrt(u)$
$\frac{d y}{d u} = \frac{1}{2 \sqrt{u}}$ $dy/(du) = 1/(2sqrt(u))$
Substituting back for $u$ $u$ we get
$\frac{d y}{d u} = \frac{1}{2 \sqrt{1 - \frac{1}{x}}}$ $dy/(du) = 1/(2sqrt(1-1/x))$

Now we shall find our $\frac{d u}{d x}$ $(du)/dx$

$u = 1 - \frac{1}{x}$ $u=1-1/x$
$\frac{d u}{d x} = 0 - (- \frac{1}{x^{2}})$ $(du)/dx = 0- (-1/x^2)$
$\frac{d u}{d x} = \frac{1}{x^{2}}$ $(du)/dx = 1/x^2$

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $dy/dx = dy/(du) xx (du)/dx$
$\frac{d y}{d x} = \frac{1}{2 \sqrt{1 - \frac{1}{x}}} \times \frac{1}{x^{2}}$ $dy/dx = 1/(2sqrt(1-1/x)) xx 1/x^2$
$\frac{d y}{d x} = \frac{1}{2 \sqrt{\frac{x - 1}{x}}} \times \frac{1}{x^{2}}$ $dy/dx = 1/(2sqrt((x-1)/x)) xx 1/x^2$

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

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What is the derivative of $\frac{\sqrt{x - 1}}{\sqrt{x}}$ $sqrt(x - 1)/sqrtx$ ?

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