View Full Version : Finding the derivative of 1/square root of X with definition
smc123
Mar 24, 2010, 05:51 PM
Can any one help me with finding the derivative of 1/x^(1/2) or square root of X using the definition F(x+h)-F(x)/h
I tried conjugating, using negative exponents but I only get really close.
Using the power rule F'(x) is clearly -1/2x^(3/2) but using the definition is more difficult
ebaines
Mar 25, 2010, 09:18 AM
Try this (for ease of typing I'm leaving the limit symbol):
Multiply through by \sqrt x \sqrt {x+h} :
\frac { \frac 1 {sqrt{ x+h}} - \frac 1 {\sqrt x}} h \times \frac {\sqrt x \sqrt {x+h} } {\sqrt x \sqrt {x+h}} = \frac { \sqrt x - \sqrt{x+h}} {h \sqrt x \sqrt {x+h} }
Whenever you have the difference of square roots like this it helps to multiply the numerator and denominator by the sum of those roots, and see where it takes you:
\frac { \sqrt x - \sqrt{x+h}} {h \sqrt x \sqrt {x+h} } =
\frac { \sqrt x - \sqrt{x+h}} {h \sqrt x \sqrt {x+h} } \cdot \frac { \ sqrt x + \sqrt{x+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})}
Now take the limit as h goes to 0.