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Math Help - 4.0 Student Stumped on simple Chain Rule App.-Please HELP!

  1. #1
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    Hello, we have just begun using the chain rule in Calculus 1, but I am stumped on a problem which is blinding me to everything else.

    It seems like it should be simple, but I just cannot get it worked out properly!
    Given: y=f(x) with f(1)=4 and f'(1)=3, find:
    - (a.) g'(1) if g(x)=sq. rt. of f(x)
    and then
    + (a.) g'(1) if g(x)=sq. rt. of f(x)....which works out to g'(1)=3/4
    but I cannot figure out:
    (b.) h'(1) if h(x)=f(sq.rt. of x)

    Please, if someone could please offer me assistance, even some hints would be appreciated.
    These are driving me insane with frustration and I have a 105% average in my class!!! What am I overlooking????


    Thank you,
    Lynn
    Last edited by mr fantastic; December 8th 2008 at 02:33 PM. Reason: Question restored by Mr F. The OP deleted by editing and cited 'personal' reasons.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by sidhlyn View Post
    Hello, we have just begun using the chain rule in Calculus 1, but I am stumped on a problem which is blinding me to everything else.

    It seems like it should be simple, but I just cannot get it worked out properly!
    Given: y=f(x) with f(1)=4 and f'(1)=3, find:
    (a.) g'(1) if g(x)=sq. rt. of f(x)
    and then
    (b.) h'(1) if h(x)=f(sq.rt. of x)

    Please, if someone could please offer me assistance, even some hints would be appreciated.
    These are driving me insane with frustration and I have a 105% average in my class!!! What am I overlooking????


    Thank you,
    Lynn
    Note that \frac{d}{dx}\left[\left(f(x)\right)^{\frac{1}{2}}\right]=\frac{1}{2}\left(f(x)\right)^{-\frac{1}{2}}\cdot f'(x)

    Do you think you can tackle part (a) now?

    Note that \frac{d}{dx}\left[f(u)\right]=f'(u)\cdot \frac{\,du}{\,dx}

    Do you think you can tackle part (b) now?

    --Chris
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  3. #3
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    Still stumped on b

    Thanks Chris...
    Actually I did manage to get (a) on my own after reviewing my notes again.
    I was just editing the post as I was sitting here pondering over it that I got part a, & saw your note after it updated.
    However, no....now I am still stumped on how to apply "u" (we use dy/du for the "inside") to part b. I'm not seeing how to apply it to f(x^1/2). ???

    Thanks again,
    Lynn
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by sidhlyn View Post
    Thanks Chris...
    Actually I did manage to get (a) on my own after reviewing my notes again.
    I was just editing the post as I was sitting here pondering over it that I got part a, & saw your note after it updated.
    However, no....now I am still stumped on how to apply "u" (we use dy/du for the "inside") to part b. I'm not seeing how to apply it to f(x^1/2). ???

    Thanks again,
    Lynn
    Let u=x^{\frac{1}{2}}

    Thus, substituting this all into the formula I gave you, we see that \frac{d}{\,dx}\left[f(u)\right]=f'(u)\cdot\frac{\,du}{\,dx}\implies\frac{d}{\,dx}  \left[f(x^{\frac{1}{2}})\right]=f'(x^{\frac{1}{2}})\cdot\frac{\,d}{\,dx}\left[x^{\frac{1}{2}}\right]=f'(x^{\frac{1}{2}})\cdot\frac{1}{2\cdot x^{\frac{1}{2}}}

    Since h'(x)=f'(x^{\frac{1}{2}})\cdot\frac{1}{2\cdot x^{\frac{1}{2}}}, what is h'(1)??

    --Chris
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  5. #5
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    Thank you, Chris!

    Oh?
    I didn't think I could do that, since the exponent affects the input itself.
    I'll give it a try.

    Thanks again!!!

    ~Lynn
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  6. #6
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    Quote Originally Posted by sidhlyn View Post
    Hello, we have just begun using the chain rule in Calculus 1, but I am stumped on a problem which is blinding me to everything else.

    It seems like it should be simple, but I just cannot get it worked out properly!
    Given: y=f(x) with f(1)=4 and f'(1)=3, find:
    - (a.) g'(1) if g(x)=sq. rt. of f(x)
    and then
    + (a.) g'(1) if g(x)=sq. rt. of f(x)....which works out to g'(1)=3/4
    but I cannot figure out:
    (b.) h'(1) if h(x)=f(sq.rt. of x)

    Please, if someone could please offer me assistance, even some hints would be appreciated.
    These are driving me insane with frustration and I have a 105% average in my class!!! What am I overlooking????


    Thank you,
    Lynn
    Thread closed due to OP deleting original question.
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