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Math Help - differentiation applying two rules

  1. #16
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    Re: differentiation applying two rules

    oh ! so is my answer incorrect ?
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  2. #17
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    Re: differentiation applying two rules

    here is my second example - please see if my inital steps are correct- while I get my head around what I did wrong in first example- ta
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  3. #18
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    Re: differentiation applying two rules

    what you mean?
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  4. #19
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    Re: differentiation applying two rules

    Ok I read what you said and I tried again- you can't fail me for trying. Please have a look
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  5. #20
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    Re: differentiation applying two rules

    Quote Originally Posted by vidhi96 View Post
    what you mean?
    For that last example, you overlooked squaring the entire denominator.

    What I meant earlier is that you are comfortable with the situation
    where the "function nesting" is one level deep, when applying the chain rule.

    e^{6x} is one level beyond e^x

    q=6x

    \frac{d}{dx}e^{6x}=\frac{d}{dx}e^q=\frac{dq}{dx} \frac{d}{dq}e^q

    However...

    v=\left(1+3x^2\right)^{\frac{1}{2}}

    has an extra stage.

    Beginning with x, we have

    w=1+3x^2

    but then we take the square root of this

    v=w^{\frac{1}{2}}

    so the "nesting" is 2 levels deep, whereas it was only 1 level deep for e^{6x}

    Hence

    \frac{d}{dx}\left(1+3x^2\right)^{\frac{1}{2}}= \frac{d}{dx}w^{\frac{1}{2}}=\frac{dw}{dx}\frac{d}{  dw}w^{\frac{1}{2}}

    =\frac{dw}{dx}\left[\frac{1}{2}w^{-\frac{1}{2}}\right]

    How is that ?
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  6. #21
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    Re: differentiation applying two rules

    Quote Originally Posted by vidhi96 View Post
    Ok I read what you said and I tried again- you can't fail me for trying. Please have a look
    On your line beginning "Therefore", you do not have the correct derivative for "v",
    but you corrected yourself on the final line!
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  7. #22
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    Re: differentiation applying two rules

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*}y &= e^{6x}(1 + 3x^2)^{\frac{1}{2}} \\ \frac{dy}{dx} &= e^{6x}\frac{d}{dx}\left[(1 + 3x^2)^{\frac{1}{2}}\right] + (1 + 3x^2)^{\frac{1}{2}}\frac{d}{dx}\left(e^{6x}\right) \end{align*}

    Now apply the chain rule to find these derivatives.
    Hey following some discussion with someone else and you- I have done some work- do you mind seeing what I have done is correct so I know I am ok to work with product and quotient rule both
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  8. #23
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    Re: differentiation applying two rules

    so you are telling me I am finally right
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  9. #24
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    Re: differentiation applying two rules

    Quote Originally Posted by Archie Meade View Post
    On your line beginning "Therefore", you do not have the correct derivative for "v",
    but you corrected yourself on the final line!
    I take it I finally understand how it works- Please say yes!!! I have also completed my own second eaxmple - heres my final version please look at it - I am typing so i can send it-therefore please excaue miner mistakes but please point to it
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  10. #25
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    Re: differentiation applying two rules

    Quote Originally Posted by vidhi96 View Post
    so you are telling me I am finally right
    Yes, but do look at why you wrote the derivative of "v" incorrectly,
    before plugging the correct derivative of v into the Product Rule formula.
    For your Quotient Rule example, you still have not squared the denominator.
    Last edited by Archie Meade; June 28th 2011 at 10:30 AM.
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  11. #26
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    Red face Re: differentiation applying two rules

    Quote Originally Posted by Archie Meade View Post
    Yes, but do look at why you wrote the derivative of "v" incorrectly,
    before plugging the correct derivative of v into the Product Rule formula.
    For your Quotient Rule example, you still have not squared the denominator.
    Ok I will do squaring and see if I can simplify in any way. However I take it rest of is ok. Is that correct? Also I know why I made mistake in "v" because I wasn't careful but since you pointed it out I think I will be extra carful in writing my steps out. Just away from the computer for next hour so when get back will send you my correction. I must say you have given me that confidence I thought I'll never have . Sincere thank you
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