# The derivative of 1/sqrt(x)

by Pithikos
Tags: advanced calculus, division, limit definition, square
 P: 48 1. The problem statement, all variables and given/known data Find the derivative of $$\frac{1}{\sqrt{x}}$$ using the lim definition. 2. Relevant equations f(x)'=$$\frac{f(x+h)-f(x)}{h}$$ 3. The attempt at a solution Keep in mind that everything bellow is for the lim as h approaches 0. $$\frac{1}{\sqrt{x}}$$ $$\Downarrow$$ $$\frac{ \frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}} } {h}$$ $$\Downarrow$$ (I multiply both nominator and denominator with conjugate) $$\frac { \frac{1}{x+h}-\frac{1}{x} } { \frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}} }$$ After this I am totally lost..
 Sci Advisor HW Helper Thanks P: 25,170 Combine numerator into a single fraction. See if you get an h you can cancel with the h in the denominator.
 P: 851 You could also use the definition... $$f'(x)= \lim_{ x \to a} \frac{f(x)- f(a)}{x-a}$$.
P: 48

## The derivative of 1/sqrt(x)

Thaaaank you! Problem solved! :)
Did the same to denominator and then combined the two franctions into one.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,882 Glad you got it solved. As a check, remember that you can write $\frac{1}{\sqrt{x}}$ as $x^{-1/2}$ and use the power rule.
Emeritus
HW Helper
PF Gold
P: 7,402
 Quote by Pithikos 1. The problem statement, all variables and given/known data Find the derivative of $$\frac{1}{\sqrt{x}}$$ using the lim definition. 2. Relevant equations f(x)'=$$\frac{f(x+h)-f(x)}{h}$$ 3. The attempt at a solution Keep in mind that everything bellow is for the lim as h approaches 0. $$\frac{1}{\sqrt{x}}$$ $$\Downarrow$$ $$\frac{ \frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}} } {h}$$ $$\Downarrow$$ (I multiply both nominator and denominator with conjugate) $$\frac { \frac{1}{x+h}-\frac{1}{x} } { \frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}} }$$ After this I am totally lost..
This is perfectly fine - up to this point.
Continuing on:

$$\displaystyle =\frac { \displaystyle \frac{x-(x+h)}{(x+h)x} } { \displaystyle \frac{h}{\ \sqrt{x+h}}+\frac{h}{\sqrt{x}\ \ } }$$

$$\displaystyle =\frac { \displaystyle \frac{-h}{(x+h)(x)}\ \cdot\ \displaystyle \frac{1}{h} } { \displaystyle \left(\frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}}\right) \ \cdot\ \displaystyle \frac{1}{h}} }$$

$$\displaystyle =\frac { \displaystyle \frac{-1}{(x+h)(x)} } { \displaystyle \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}} }$$

Then,
$$\displaystyle f'(x)= \lim_{h\to 0} \ \ \frac { \displaystyle \frac{-1}{(x+h)(x)} } { \displaystyle \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}} }$$

 Related Discussions Calculus & Beyond Homework 11 Calculus & Beyond Homework 1 Calculus & Beyond Homework 1 Calculus & Beyond Homework 1 General Math 5