Newton's Difference Quotient :D

[stinajeana]

Let f(x)=1/(sqrt x). Use Newton's Difference Quotient to prove that f'(x)= -1/(2xsqrtx)
...in case the last part doesn't make sense this is what it is in words "negative one divided by two x square root x" such that the 2x is right beside the square root of x and not in the index

How do I start?

I was thinking that I should plug the 1/(sqrt x) where "f" is in Newton's Difference Quotient ... but i'm not sure what the do after that.

 

[daon2]

Let f(x)=1/(sqrt x). Use Newton's Difference Quotient to prove that f'(x)= -1/(2xsqrtx)
...in case the last part doesn't make sense this is what it is in words "negative one divided by two x square root x" such that the 2x is right beside the square root of x and not in the index

How do I start?

I was thinking that I should plug the 1/(sqrt x) where "f" is in Newton's Difference Quotient ... but i'm not sure what the do after that.

\(\displaystyle \displaystyle \Delta f = f(x+\Delta x) - f(x)=\dfrac{1}{\sqrt{x+\Delta x}}-\dfrac{1}{\sqrt{x}}\)

You want to calculate:

\(\displaystyle \displaystyle \lim_{\Delta x\to 0} \dfrac{\Delta f}{\Delta x} = \lim_{\Delta x\to 0}\dfrac{\dfrac{1}{\sqrt{x+\Delta x}}-\dfrac{1}{\sqrt{x}}}{\Delta x}\)

To start, multiply the entire fraction by "1" in the form of \(\displaystyle \dfrac{\sqrt{x}\sqrt{x+\Delta x}}{\sqrt{x}\sqrt{x+\Delta x}}\). Distribute the top of this to the numerator to your difference quotient and simplify, and chuck the rest into the bottom with the delta x. The denominator will remain "ugly" for the rest of the problem. But you can apply another technique to make the numerator prettier.

 

[JeffM]

Let f(x)=1/(sqrt x). Use Newton's Difference Quotient to prove that f'(x)= -1/(2xsqrtx)
...in case the last part doesn't make sense this is what it is in words "negative one divided by two x square root x" such that the 2x is right beside the square root of x and not in the index

How do I start?

I was thinking that I should plug the 1/(sqrt x) where "f" is in Newton's Difference Quotient ... but i'm not sure what the do after that.

\(\displaystyle f(x) = \dfrac{1}{\sqrt{x}} \implies \)

\(\displaystyle f(x + h) = \dfrac{1}{\sqrt{x + h}} \implies\)

\(\displaystyle f(x + h) - f(x) = \dfrac{1}{\sqrt{x + h}} - \dfrac{1}{\sqrt{x}} = \dfrac{\sqrt{x}}{\sqrt{x + h} * \sqrt{x}} - \dfrac{\sqrt{x + h}}{\sqrt{x} * \sqrt{x + h}} = \dfrac{\sqrt{x} - \sqrt{x + h}}{\sqrt{x^2 + hx}} \implies\)

\(\displaystyle \dfrac{f(x + h) - f(x)}{h} = \left(\dfrac{\sqrt{x} -\sqrt{x + h}}{\sqrt{x^2 + hx}}\right) \div h = \dfrac{\sqrt{x} -\sqrt{x + h}}{h\sqrt{x^2 + hx}} = \dfrac{\sqrt{x} -\sqrt{x + h}}{h\sqrt{x^2 + hx}} * \dfrac{\sqrt{x} +\sqrt{x + h}}{\sqrt{x} +\sqrt{x + h}}.\)

Now finish it up.

 

[stinajeana]

[QUOTE Now finish it up.[/QUOTE]

Is -1/(sqrtx)(sqrtx)*(sqrtx + sqrtx) the same as -1/(2xsqrtx)?

 

[lookagain]

[QUOTE Now finish it up.

Is -1/(sqrtx)(sqrtx)*(sqrtx + sqrtx) the same as -1/(2xsqrtx)?[/QUOTE]
No. You must have grouping symbols, such as the following: \(\displaystyle \ \ \ \)-1/[sqrt(x)*sqrt(x)*(sqrt(x) + sqrt(x))] = -1/[x*2sqrt(x)] = -1/[2xsqrt(x)]

 

[stinajeana]

Is -1/(sqrtx)(sqrtx)*(sqrtx + sqrtx) the same as -1/(2xsqrtx)?

No. You must have grouping symbols, such as the following: \(\displaystyle \ \ \ \)-1/[sqrt(x)*sqrt(x)*(sqrt(x) + sqrt(x))] = -1/[x*2sqrt(x)] = -1/[2xsqrt(x)][/QUOTE]

Thank you very much for your help!