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Math Help - differentiation applying two rules

  1. #1
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    differentiation applying two rules

    Hi I am fine when it comes to straightforward differntiation but when its composite I mean when you have to apply say product rule and then chain rule etc. I really haven't got a clue when to proceed with second rule. I have a problem here and I will really appriciate it if someone can explain in details:

    f(x)=e^(6x) sq.rt.(1+3x^2) I don't know how to use fancy text but this should read
    f(x) = e to the power of 6x multiplied by square root of 1 plus 3x squared.

    I know you apply prodcut and then composite rule- but I get stuck as soon as I did prodcut rule. if someone can illustarte the whole process and logic that will make my life easier
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    Re: differentiation applying two rules

    Quote Originally Posted by vidhi96 View Post
    Hi I am fine when it comes to straightforward differntiation but when its composite I mean when you have to apply say product rule and then chain rule etc. I really haven't got a clue when to proceed with second rule. I have a problem here and I will really appriciate it if someone can explain in details:

    f(x)=e^(6x) sq.rt.(1+3x^2) I don't know how to use fancy text but this should read
    f(x) = e to the power of 6x multiplied by square root of 1 plus 3x squared.

    I know you apply prodcut and then composite rule- but I get stuck as soon as I did prodcut rule. if someone can illustarte the whole process and logic that will make my life easier
    The Product Rule is

    (uv)'=u'(v)+v'(u)

    Hence you write your derivatives for u and v
    and plug them into the formula.

    This will be relatively easy if you can get those derivatives,
    which require the Chain Rule.

    How is this so far ?
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  3. #3
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    Re: differentiation applying two rules

    \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.
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  4. #4
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    Re: differentiation applying two rules

    This is where I get stuck- so I did prodcut rule and got as far as you have - now when you apply chain rule- which part do you aplly it to as I am thorughly confused.


    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.
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  5. #5
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    Re: differentiation applying two rules

    Hi Please have a look at attachment of what I have done



    Quote Originally Posted by Archie Meade View Post
    The Product Rule is

    (uv)'=u'(v)+v'(u)

    Hence you write your derivatives for u and v
    and plug them into the formula.

    This will be relatively easy if you can get those derivatives,
    which require the Chain Rule.

    How is this so far ?
    Attached Files Attached Files
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  6. #6
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    Re: differentiation applying two rules

    please have a look at what I have done and tell me if my understanding is correct or not Thanks
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  7. #7
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    Re: differentiation applying two rules

    Wasn't able to open that, but...

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    But this is wrapped inside the right hand leg of the legs-uncrossed version of...



    ... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.



    Spoiler:



    Or, zooming in on the exponential also...





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  8. #8
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    Re: differentiation applying two rules

    hey Guys is this the right answer
    dy/dx=6e^6x 〖(1+3x^2)〗^(1/2)+e^6x 1/2 〖(1+3x^2)〗^(-1/2)
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  9. #9
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    Re: differentiation applying two rules

    ok heres a PDF hope you can open it- while I try and figure it out your diagram. I am not good at this cal but I must learn for my exam- so every help is much appriciated
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  10. #10
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    Re: differentiation applying two rules

    Quote Originally Posted by vidhi96 View Post
    hey Guys is this the right answer
    dy/dx=6e^6x 〖(1+3x^2)〗^(1/2)+e^6x 1/2 〖(1+3x^2)〗^(-1/2)
    What you're missing there is the 6x in the bottom-right corner of my diagrams. (Arising from the chain rule for differentiation, just as the 6 popped out from inside the exponential - you included that one, just do the same on the right, as per the diagram.)

    PS the third (or fourth) balloon along has the sqrt tweaked into a fraction to prepare for simplifying.
    Last edited by tom@ballooncalculus; June 28th 2011 at 08:56 AM.
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  11. #11
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    Re: differentiation applying two rules

    sorry I have self corrected myself and come to this final answer
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  12. #12
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    Re: differentiation applying two rules

    hey please see my correction thanks
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  13. #13
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    Re: differentiation applying two rules

    Quote Originally Posted by vidhi96 View Post
    ok heres a PDF hope you can open it- while I try and figure it out your diagram. I am not good at this cal but I must learn for my exam- so every help is much appriciated
    You just need a bit more co-ordination.

    On the attachment, you wrote

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

    This is in fact "v" written in "index form" in preparation for differentiation.

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

    You correctly differentiated e^{6x}
    using the Chain Rule.

    e^{6x}=e^u

    \frac{d}{dx}e^u=\frac{du}{dx}\frac{d}{du}e^u

    To differentiate "v"

    w=1+3x^2=f(x)

    v=w^{\frac{1}{2}}=f(w)

    \frac{dv}{dx}=\frac{dv}{dw}\frac{dw}{dx}

    Writing the stages in this way helps simplify the Chain Rule calculations.
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  14. #14
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    Re: differentiation applying two rules

    i am bit confident- I am going to work through another example and please guide me so I know I can do this. I will attach what I have done soon for you to look at thanks again
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  15. #15
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    Re: differentiation applying two rules

    Quote Originally Posted by vidhi96 View Post
    i am bit confident- I am going to work through another example and please guide me so I know I can do this. I will attach what I have done soon for you to look at thanks again
    You will have it when you've mastered differentiating the "v" part.
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