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Math Help - The Chain Rule

  1. #1
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    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!!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    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 f(x)=x^2\sqrt{9-x^2}

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

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

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

    So we see that 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}} =-\frac{x}{\sqrt{9-x^2}}

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

    If f(x)=u(x)v(x), then 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; October 1st 2008 at 08:14 PM. Reason: typo
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  3. #3
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    For the first one put f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6}, hence 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}}}.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Krizalid View Post
    For the first one put f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6}, hence 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
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  5. #5
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    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??
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    For the first one put f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6}, hence 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!
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    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:

    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] 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 \sqrt{9-x^2}.

    So we see that 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}} =\frac{18x-2x^3-x^3}{\sqrt{9-x^2}}=\color{red}\boxed{\frac{18x-3x^3}{\sqrt{9-x^2}}}
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  8. #8
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    NOW I understand the problem.. Thanks!
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