need help finding the derivative of f(x)=1/[x^(1/2)]

[samantha0417]

I need to find the derivative for the following problem:

f(x)=1/[x^(1/2)]

**The denominator is the square root of x**

I have to find the drivative using [f(x+h) - f(x)]/h

I have worked this problem out several times and each time I have a "h" left on the denominator that will cause the solution to be underfined.

 

[stapel]

I have worked this problem out several times and....

Great! Please reply showing those steps, so the tutors can help you find any errors, or else confirm your work and provide direction on the next step(s) to take.

Please be specific. Thank you.

Eliz.

 

[samantha0417]

f'(x)= lim as x goes to 0 [x^(.5)-(x+h)^(.5)]/[h(x^(.5)*(x+h)^(.5))]
then I canceled out the "x"s on the numerator to get

-h^.5/[h*x^(.5)(x+h)^.5]

 

[soroban]

Hello, Samantha!

Find the derivative of: \(\displaystyle \,f(x)\:=\:\L\frac{1}{\sqrt{x}}\)

using: \(\displaystyle \,\lim_{h\to0}\frac{f(x\,+\,h)\,-\,f(x)}{h}\)

The numerator is: \(\displaystyle \,f(x\,+\,h)\,-\,f(x)\;=\;\L\frac{1}{\sqrt{x\,+\,h}}\,-\,\frac{1}{\sqrt{x}}\)

Combine: \(\displaystyle \L\,\frac{1}{\sqrt{x\,+\,h}}\cdot\frac{\sqrt{x}}{\sqrt{x}}\,-\,\frac{1}{\sqrt{x}}\cdot\frac{\sqrt{x\,+\,h}}{\sqrt{x\,+\,h}} \;= \;\frac{\sqrt{x} \,- \,\sqrt{x\,+\,h}}{\sqrt{x}\cdot\sqrt{x\,+\,h}}\)

Rationalize: \(\displaystyle \L\:\frac{\sqrt{x}\,-\,\sqrt{x\,+\,h}}{\sqrt{x}\cdot\sqrt{x\,+\,h}}\cdot\frac{\sqrt{x}\,+\,\sqrt{x\,+\,h}}{\sqrt{x}\,+\,\sqrt{x\,+\,h}} \;= \;\frac{x\,-\,(x\,+\,h)}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)}\)

. . and we have: \(\displaystyle \L\:\frac{-h}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)}\)

Divide by \(\displaystyle h:\L\;\;\frac{1}{\not{h}}\cdot\frac{-\not{h}}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)} \;=\;\frac{-1}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)}\)

Take the limit: \(\displaystyle \L\,f'(x) \;= \;\lim_{h\to0}\,\frac{-1}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)} \;=\;\frac{-1}{\sqrt{x}\cdot\sqrt{x}\left(\sqrt{x}\,+\,\sqrt{x}\right)}\)

Therefore: \(\displaystyle \L\,f'(x)\;=\;\frac{-1}{x\cdot2\sqrt{x}} \;=\;-\frac{1}{2x^{\frac{3}{2}}}\)