Results 1 to 5 of 5

Math Help - Find y prime

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    118

    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?)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Find y prime

    Quote Originally Posted by sparky View Post
    \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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1

    Re: Find y prime

    Quote Originally Posted by sparky View Post
    \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})}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2009
    Posts
    118

    Re: Find y prime

    Thanks for your reply chisigma.

    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.
    Last edited by sparky; September 10th 2011 at 07:35 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428

    Re: Find y prime

    Quote Originally Posted by sparky View Post
    \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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 3rd 2011, 06:49 PM
  2. Find derivative. or Y prime.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 25th 2010, 07:54 AM
  3. find all prime numbers
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 16th 2009, 02:38 PM
  4. Find all prime numbers
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: August 22nd 2008, 04:48 AM
  5. Find a prime number
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: January 31st 2008, 07:47 PM

Search Tags


/mathhelpforum @mathhelpforum