# Finding the derivative using the definition (have answer)

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• February 26th 2010, 09:42 AM
janedoe
Finding the derivative using the definition (have answer)
Find the derivative using the definition (NOT shortcuts, we must use definition).

y=1/sqrt (x)

My friend and I got it down to x^(-1) - x^(-3/2)
but the answer is: -1/2x^(3/2)

How did they get that? (not using shortcuts)
Thanks!
• February 26th 2010, 10:02 AM
Plato
$\frac{{\frac{1}
{{\sqrt {x + h} }} - \frac{1}
{{\sqrt x }}}}
{h} = \frac{{\sqrt x - \sqrt {x + h} }}
{{h\sqrt x \left( {\sqrt {x + h} } \right)}} = \frac{{ - h}}
{{h\sqrt x \left( {\sqrt {x + h} } \right)\left( {\sqrt x + \sqrt {x + h} } \right)}}$
• February 26th 2010, 10:09 AM
Random Variable
$\lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x}$

$= \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x} \cdot \frac{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}$

$= \lim_{\Delta x \to 0} \frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x \Big(\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}\Big)}$

$= \lim_{\Delta x \to 0} \frac{\frac{-1}{x(x+ \Delta x)}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}$

$= \frac{\frac{-1}{x^{2}}}{\frac{2}{\sqrt{x}}} = \frac{-1}{2x^{3/2}}$
• February 26th 2010, 10:29 AM
janedoe
Quote:

Originally Posted by Random Variable
$\lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x}$

$= \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x} \cdot \frac{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}$

$= \lim_{\Delta x \to 0} \frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x \Big(\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}\Big)}$

$= \lim_{\Delta x \to 0} \frac{\frac{-1}{x(x+ \Delta x)}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}$

$= \frac{\frac{-1}{x^{2}}}{\frac{2}{\sqrt{x}}} = \frac{-1}{2x^{3/2}}$

THANK YOU!! You're a genius lol. I'm so happy I know how to do this now. :]