# Thread: derivative of 1/[sqrt[x]: proving with derivative definition

1. ## derivative of 1/[sqrt[x]: proving with derivative definition

i need to solve 1/sqrt[x] using the limit method. i checked it first using the various rules and got 1/2x^3/2. heres what i did:

placed the function in the definition

1/sqrt[(x+h)] - 1/sqrt[x]
----------------------------
h

[(x+h)^-.5] - [x^-.5]
------------------------
h

apply conjugate:
[(x+h)^-.5] - [x^-.5] * [(x+h)^-.5] + [x^-.5]
------------------------
h * [(x+h)^-.5] + [x^-.5]

im lost here on out.

2. Originally Posted by Evan.Kimia
i need to solve 1/sqrt[x] using the limit method. i checked it first using the various rules and got 1/2x^3/2. heres what i did:

placed the function in the definition

1/sqrt[(x+h)] - 1/sqrt[x]
----------------------------
h

[(x+h)^-.5] - [x^-.5]
------------------------
h

apply conjugate:
[(x+h)^-.5] - [x^-.5] * [(x+h)^-.5] + [x^-.5]
------------------------
h * [(x+h)^-.5] + [x^-.5]

im lost here on out.
$\frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} = \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} = \frac{x-x-h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}$ $= \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} \to \frac{-1}{\sqrt{x}\sqrt{x}(\sqrt{x} + \sqrt{x})} = \frac{-1}{2x^{\frac{3}{2}}}$

3. Originally Posted by Evan.Kimia
i need to solve 1/sqrt[x] using the limit method. i checked it first using the various rules and got 1/2x^3/2. heres what i did:

placed the function in the definition

1/sqrt[(x+h)] - 1/sqrt[x]
----------------------------
h

[(x+h)^-.5] - [x^-.5]
------------------------
h

apply conjugate:
[(x+h)^-.5] - [x^-.5] * [(x+h)^-.5] + [x^-.5]
------------------------
h * [(x+h)^-.5] + [x^-.5]

im lost here on out.

Good idea to use the conjugate but you have to push forward a little more:

$\lim_{h\to 0}\,\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\lim_{h\to 0}\,\frac{\frac{1}{x+h}-\frac{1}{x}}{h\left(\frac{1}{\sqrt{x+h}}+\frac{1}{ \sqrt{x}}\right)}$ $=\lim_{h\to 0}\,\frac{-h}{h\left(\frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}\ right)x(x+h)}=\lim_{h\to 0}\,\frac{-1}{\left(\frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}\r ight)x(x+h)}$ $=-\frac{1}{\frac{2}{\sqrt{x}}x^2}=-\frac{1}{2x^{3/2}}=-\frac{1}{2x\sqrt{x}}$

Tonio

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