# Math Help - Find y prime

1. ## Find y prime

$\text{Find } y' (1) \text{if } y(x) = \frac{1}{\sqrt x}$

Hi, I need help completing the question above. Here is what I have so far:

$= \frac{y(1+h) - y(1)}{h}$

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

Am I correct so far? What do I do next? (and why do I do what I do next?)

2. ## Re: Find y prime

Originally Posted by sparky
$\text{Find } y' (1) \text{if } y(x) = \frac{1}{\sqrt x}$

Hi, I need help completing the question above. Here is what I have so far:

$= \frac{y(1+h) - y(1)}{h}$

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

Am I correct so far? What do I do next? (and why do I do what I do next?)
Why don't 'attack' the general problem: what's the derivative of $x^{\alpha}$?... take into account that is $x^{\alpha}= e^{\alpha \ln x}$ and that You can always use the product rule for derivatives...

Kind regards

$\chi$ $\sigma$

3. ## Re: Find y prime

Originally Posted by sparky
$\text{Find } y' (1) \text{if } y(x) = \frac{1}{\sqrt x}$

Hi, I need help completing the question above. Here is what I have so far:

$= \frac{y(1+h) - y(1)}{h}$

$= (\frac{1}{\sqrt (1+h)} - \frac{1}{\sqrt 1}) * \frac{1}{h}$
Note that $= (\frac{1}{\sqrt (1+h)} - \frac{1}{\sqrt 1})=\frac{-h}{\sqrt{1+h}(1+\sqrt{1+h})}$

4. ## Re: Find y prime

My teacher used the difference quotient to work this question out but I did not understand beyond what I wrote above. I am trying to understand this using the difference quotient.

Derivative of $x^\alpha$? I will do some research on this because I have no idea what this is about.

5. ## Re: Find y prime

Originally Posted by sparky
$\text{Find } y' (1) \text{if } y(x) = \frac{1}{\sqrt x}$

Hi, I need help completing the question above. Here is what I have so far:

$= \frac{y(1+h) - y(1)}{h}$

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

Am I correct so far? What do I do next? (and why do I do what I do next?)
common denominator ...

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

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

rationalize the numerator ...

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

$\left[\frac{1-(1+h)}{\sqrt{1+h}(1+\sqrt{1+h})}\right] \cdot \frac{1}{h}$

combine like terms and divide out the resulting factor $h$ in the numerator ...

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

now take the limit as $h \to 0$