Results 1 to 8 of 8

Thread: The Chain Rule

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    126

    The Chain Rule

    Hey.. I'm having some serious problems with the Chain Rule! It's my 2nd day on it and I'm getting a little frustrated on these extra problems I am doing but can't get right...

    1. x^2√(9-x^2)
    I know this is the product rule but I don't really get it.. Whenever I use the product rule then take the derivative of √(9-x^2) I just get LOST!!!

    2. secx^2

    I always get the wrong answer everytime and can't figure it out..

    h'(x)= secxtanx X 2(secx)
    = 2sec^2xtanx

    The answer shows up as 2xsecx^2tanx^2

    help!! I need help understanding the chain rule and I got no idea how to do problems like these!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by elpermic View Post
    Hey.. I'm having some serious problems with the Chain Rule! It's my 2nd day on it and I'm getting a little frustrated on these extra problems I am doing but can't get right...

    1. x^2√(9-x^2)
    I know this is the product rule but I don't really get it.. Whenever I use the product rule then take the derivative of √(9-x^2) I just get LOST!!!
    If you're new to this, I would approach it this way.

    We have $\displaystyle f(x)=x^2\sqrt{9-x^2}$

    Let $\displaystyle u(x)=x^2$ and $\displaystyle v(x)=\sqrt{9-x^2}$

    We see that $\displaystyle u'(x)=2x$ and to find $\displaystyle v'(x)$, apply the chain rule.

    Note that the chain rule is $\displaystyle \frac{d}{\,dx}\left[f(g(x))\right]=f'(g(x))\cdot g'(x)$.

    So we see that $\displaystyle v'(x)=\tfrac{1}{2}(9-x^2)^{-\frac{1}{2}}\cdot\frac{d}{\,dx}\left[9-x^2\right]\implies v'(x)=\frac{1}{2\sqrt{9-x^2}}\cdot(-2x)=-\frac{2x}{2\sqrt{9-x^2}}$ $\displaystyle =-\frac{x}{\sqrt{9-x^2}}$

    Now substitute $\displaystyle u(x),~v(x),~u'(x),~\text{and}~v'(x)$ into the equation for the product rule:

    If $\displaystyle f(x)=u(x)v(x)$, then $\displaystyle f'(x)=u(x)v'(x)+v(x)u'(x)$

    Can you take it from here?

    2. secx^2

    I always get the wrong answer everytime and can't figure it out..

    h'(x)= secxtanx X 2(secx)
    = 2sec^2xtanx

    The answer shows up as 2xsecx^2tanx^2

    help!! I need help understanding the chain rule and I got no idea how to do problems like these!!
    I don't see anything wrong with your answer. Make sure you copied down the problem correctly...otherwise, I'd have to say the book's answer is wrong.

    --Chris
    Last edited by Chris L T521; Oct 1st 2008 at 08:14 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    For the first one put $\displaystyle f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6},$ hence $\displaystyle f'(x)=\frac{1}{2\sqrt{9x^{4}-x^{6}}}\cdot \left( 9x^{4}-x^{6} \right)'=\frac{9\cdot 4x^{3}-6x^{5}}{2x^{2}\sqrt{9-x^{2}}}=\frac{18x-3x^{3}}{\sqrt{9-x^{2}}}.$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by Krizalid View Post
    For the first one put $\displaystyle f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6},$ hence $\displaystyle f'(x)=\frac{1}{2\sqrt{9x^{4}-x^{6}}}\cdot \left( 9x^{4}-x^{6} \right)'=\frac{9\cdot 4x^{3}-6x^{5}}{2x^{2}\sqrt{9-x^{2}}}=\frac{18x-3x^{3}}{\sqrt{9-x^{2}}}.$
    Ha! I was thinking about that...but he was working with product and chain rules...so I did it the way he was doing it. It is always good to see another approach

    --Chris
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2008
    Posts
    126
    I was working with your way and I still don't get how to do it...

    f'(x)= (2x)(√(9-x^2) + (-x/2√(9-x^2))(x^2)
    = 2x√(9-x^2) - (-x^3/2√(9-x^2))

    Did I do something wrong??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2008
    Posts
    126
    Quote Originally Posted by Krizalid View Post
    For the first one put $\displaystyle f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6},$ hence $\displaystyle f'(x)=\frac{1}{2\sqrt{9x^{4}-x^{6}}}\cdot \left( 9x^{4}-x^{6} \right)'=\frac{9\cdot 4x^{3}-6x^{5}}{2x^{2}\sqrt{9-x^{2}}}=\frac{18x-3x^{3}}{\sqrt{9-x^{2}}}.$
    That is the answer but how do you get it using the product&chain rule?? I've got no idea what you just did BTW!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by elpermic View Post
    I was working with your way and I still don't get how to do it...

    f'(x)= (2x)(√(9-x^2) + (-x/2√(9-x^2))(x^2)
    = 2x√(9-x^2) + (-x^3/2√(9-x^2))

    Did I do something wrong??
    The bit in red should be plus, and the 2 shouldn't be there. This is why:

    $\displaystyle f'(x)=\underbrace{(2x)}_{u'(x)}\underbrace{(\sqrt{ 9-x^2})}_{v(x)}+\underbrace{\left(-\frac{x}{\sqrt{9-x^2}}\right)}_{v'(x)}\underbrace{(x^2)}_{u(x)}=2x\ sqrt{9-x^2}+\frac{-x^3}{\sqrt{9-x^2}}$

    So we have [as of now] $\displaystyle f'(x)=2x\sqrt{9-x^2}+\frac{-x^3}{\sqrt{9-x^2}}$

    To combine the two terms, they need to have a common denominator. In this case, the common denominator is $\displaystyle \sqrt{9-x^2}$.

    So we see that $\displaystyle 2x\sqrt{9-x^2}\cdot{\color{red}\frac{\sqrt{9-x^2}}{\sqrt{9-x^2}}}+\frac{-x^3}{\sqrt{9-x^2}}=\frac{2x(9-x^2)-x^3}{\sqrt{9-x^2}}$ $\displaystyle =\frac{18x-2x^3-x^3}{\sqrt{9-x^2}}=\color{red}\boxed{\frac{18x-3x^3}{\sqrt{9-x^2}}}$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2008
    Posts
    126
    NOW I understand the problem.. Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: Nov 9th 2010, 01:40 AM
  2. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 22nd 2009, 08:50 PM
  3. Replies: 5
    Last Post: Oct 19th 2009, 01:04 PM
  4. Replies: 3
    Last Post: May 25th 2009, 06:15 AM
  5. Replies: 2
    Last Post: Dec 13th 2007, 05:14 AM

Search Tags


/mathhelpforum @mathhelpforum