How Do You Differentiate 1/sqrt(x) Using Limits?

[Pithikos]

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.

$$\frac{1}{\sqrt{x}}$$

$$\Downarrow$$

$$\frac{<br /> \frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}<br /> }<br /> {h}$$

$$\Downarrow$$

(I multiply both nominator and denominator with conjugate)

$$\frac<br /> {<br /> \frac{1}{x+h}-\frac{1}{x}<br /> }<br /> {<br /> \frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}}<br /> }$$

After this I am totally lost..

 

[Dick]

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

 

[╔(σ_σ)╝]

You could also use the definition...

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

 

[Pithikos]

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

 

[HallsofIvy]

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.

 

[SammyS]

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.

$$\frac{1}{\sqrt{x}}$$

$$\Downarrow$$

$$\frac{<br /> \frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}<br /> }<br /> {h}$$

$$\Downarrow$$

(I multiply both nominator and denominator with conjugate)

$$\frac<br /> {<br /> \frac{1}{x+h}-\frac{1}{x}<br /> }<br /> {<br /> \frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}}<br /> }$$

After this I am totally lost..

This is perfectly fine - up to this point.
Continuing on:

$$\displaystyle =\frac<br /> { \displaystyle \frac{x-(x+h)}{(x+h)x}<br /> }<br /> { \displaystyle \frac{h}{\ \sqrt{x+h}}+\frac{h}{\sqrt{x}\ \ }<br /> }$$

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

$$\displaystyle =\frac<br /> { \displaystyle \frac{-1}{(x+h)(x)}<br /> }<br /> { \displaystyle \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}<br /> }$$

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