# The derivative of 1/sqrt(x)

• Pithikos
\displaystyle = \frac{-1}{x\cdot x}\cdot \lim_{h\to 0} \ \ \frac{1}{\left(1+\frac{h}{x}\right)} \cdot \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}} } \displaystyle = \frac{-1}{x^2}\cdot \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}} \displaystyle = \frac{-1+1}{x^{2}\sqrt{x}}} \displaystyle = \frac{0}{x^{2}\sqrt{x}

## Homework Statement

Find the derivative of $$\frac{1}{\sqrt{x}}$$ using the lim definition.

## Homework Equations

f(x)'=$$\frac{f(x+h)-f(x)}{h}$$

## 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..

Combine numerator into a single fraction. See if you get an h you can cancel with the h in the denominator.

You could also use the definition...

$$f'(x)= \lim_{ x \to a} \frac{f(x)- f(a)}{x-a}$$.

Thaaaank you! Problem solved! :)
Did the same to denominator and then combined the two franctions into one.

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.

Pithikos said:

## Homework Statement

Find the derivative of $$\frac{1}{\sqrt{x}}$$ using the lim definition.

## Homework Equations

f(x)'=$$\frac{f(x+h)-f(x)}{h}$$

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