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Math Help - Confunsing Chain Rules and more

  1. #1
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    Confunsing Chain Rules and more

    I need a bit of help with my assignment it due monday but I am in utter confusement.

    Here is my first problem I need help making I did right

    someone please check this derivative for me

    the function is
    f(x)=e^{2x}ln(10x^2+3)

    f\prime(x)=(2e^{2x})[ln(10x^{2}+3)]+(e^{2x})(\frac{1}{10x^2+3})
    <br />
f\prime(x)=e^{2x}[2(ln(10x^2+3)]+(\frac{1}{10x^2+3})

    ?

    Also how would I begin to simplify something like this take the derivativea

    -\frac{x^2+x^3-\pi}{x^-3}?
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    Quote Originally Posted by sk8erboyla2004 View Post
    I need a bit of help with my assignment it due monday but I am in utter confusement.

    Here is my first problem I need help making I did right

    someone please check this derivative for me

    the function is
    f(x)=e^{2x}ln(10x^2+3)

    f\prime(x)=(2e^{2x})[ln(10x^{2}+3)]+(e^{2x})(\frac{1}{10x^2+3})
    <br />
f\prime(x)=e^{2x}[2(ln(10x^2+3)]+(\frac{1}{10x^2+3})

    ?

    Also how would I begin to simplify something like this take the derivativea

    -\frac{x^2+x^3-\pi}{x^-3}?
    If you have combinations of product, chain or quotient rules, it helps to take things one step at a time.

    So for the first...

    f = e^{2x}\ln{(10x^2 + 3)}

    Use the product rule

    \frac{df}{dx} = e^2x\frac{d}{dx}[\ln{(10x^2 + 3)}] + \ln{(10x^2 + 3)}\frac{d}{dx}(e^{2x}).

    Now use the chain rule to work out these easier derivatives.

    \frac{d}{dx}(e^{2x}) is easy, it's just 2e^{2x}.

    If y = \ln{(10x^2 + 3)} let u = 10x^2 + 3 so that y = \ln{u}.

    \frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du}.

    \frac{du}{dx} = 20x

    \frac{dy}{du} = \frac{1}{u} = \frac{1}{10x^2 + 3}.

    So \frac{dy}{dx} = \frac{20x}{10x^2 + 3}.


    Therefore

    \frac{df}{dx} = \frac{20xe^{2x}}{10x^2 + 3} + 2e^{2x}\ln{(10x^2 + 3)}.
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  3. #3
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    Quote Originally Posted by sk8erboyla2004 View Post
    I need a bit of help with my assignment it due monday but I am in utter confusement.

    Here is my first problem I need help making I did right

    someone please check this derivative for me

    the function is
    f(x)=e^{2x}ln(10x^2+3)

    f\prime(x)=(2e^{2x})[ln(10x^{2}+3)]+(e^{2x})(\frac{1}{10x^2+3})
    <br />
f\prime(x)=e^{2x}[2(ln(10x^2+3)]+(\frac{1}{10x^2+3})

    ?

    Also how would I begin to simplify something like this take the derivativea

    -\frac{x^2+x^3-\pi}{x^-3}?
    f = -\frac{x^2+x^3-\pi}{x^{-3}} is a quotient, so use the quotient rule.

    \frac{df}{dx} = -\left[\frac{x^{-3}\frac{d}{dx}(x^2 + x^3 - \pi) - (x^2 + x^3 - \pi)\frac{d}{dx}(x^{-3})}{(x^{-3})^2}\right]

     = -\left[\frac{x^{-3}(2x + 3x^2) + 3x^{-4}(x^2 + x^3 - \pi)}{x^{-6}}\right]

     = -[x^3(2x + 3x^2) + 3x^2(x^2 + x^3 - \pi)].
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  4. #4
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    Quote Originally Posted by Prove It View Post
    If you have combinations of product, chain or quotient rules, it helps to take things one step at a time.

    So for the first...

    f = e^{2x}\ln{(10x^2 + 3)}

    Use the product rule

    \frac{df}{dx} = e^2x\frac{d}{dx}[\ln{(10x^2 + 3)}] + \ln{(10x^2 + 3)}\frac{d}{dx}(e^{2x}).

    Now use the chain rule to work out these easier derivatives.

    \frac{d}{dx}(e^{2x}) is easy, it's just 2e^{2x}.

    If y = \ln{(10x^2 + 3)} let u = 10x^2 + 3 so that y = \ln{u}.

    \frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du}.

    \frac{du}{dx} = 20x

    \frac{dy}{du} = \frac{1}{u} = \frac{1}{10x^2 + 3}.

    So \frac{dy}{dx} = \frac{20x}{10x^2 + 3}.


    Therefore

    \frac{df}{dx} = \frac{20xe^{2x}}{10x^2 + 3} + 2e^{2x}\ln{(10x^2 + 3)}.
    but isnt the derivative of the natural log 1/x? so where does that come into play if it does

    and thanks!

    nvm I think I understood that part man I am sure dumbfounded as when to use the cahin rule
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    Quote Originally Posted by sk8erboyla2004 View Post
    but isnt the derivative of the natural log 1/x? so where does that come into play if it does

    and thanks!
    Yes, the derivative of a natural logarithm of the form

    y = \ln{x} is \frac{1}{x}.

    But if you have a natural logarithm of a FUNCTION of x, you have to use the chain rule.

    I've shown it to you step by step already...
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    Yes ive seen but I have no clue how to implement this chain rule properly
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    Quote Originally Posted by sk8erboyla2004 View Post
    Yes ive seen but I have no clue how to implement this chain rule properly
    Read it carefully

    If let so that .

    .



    .

    So .
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    Yes, I understand and I can clearly see in the example of that chain rule is

    f\prime(g(x))*g\prime(x)

    and that

    f\prime(g(x)=\frac{1}{10x^2+3}

    and

    g\prime(x)=20x

    What I rather meant to say is I am unclear of when to use the chain rule and the extent of using it for more than two function as per say

    Thanks though!


    check my answer for this one as well

    y=\frac{e^{\pi}}{\pi^{e}}

    y\prime=(e^{\pi})(\pi^{e}+e\pi^{e-1})
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    Quote Originally Posted by sk8erboyla2004 View Post
    Yes, I understand and I can clearly see in the example of that chain rule is

    f\prime(g(x))*g\prime(x)

    and that

    f\prime(g(x)=\frac{1}{10x^2+3}

    and

    g\prime(x)=20x

    What I rather meant to say is I am unclear of when to use the chain rule and the extent of using it for more than two function as per say

    Thanks though!


    check my answer for this one as well

    y=\frac{e^{\pi}}{\pi^{e}}

    y\prime=(e^{\pi})(\pi^{e}+e\pi^{e-1})
    Yes it can be difficult. You just need practice.

    Um, for your next one, \frac{e^\pi}{\pi^e} is a constant.

    So it's derivative is 0.
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    Quote Originally Posted by Prove It View Post
    Yes it can be difficult. You just need practice.

    Um, for your next one, \frac{e^\pi}{\pi^e} is a constant.

    So it's derivative is 0.

    Darn I need mad practice, ugh.

    Thanks also check this one it has far as i got and im unsure of how to future simply it bc i maybe be wrong

    h(x)=\frac{(2x^2-4x^6)^4}{2+2x^2}

    h\prime(x)=\frac{4(2x^2-4x^6)^3(4x-24x^5)(2+2x^2)-(4x)(2x^2-4x^6)^4}{(2+2x^2)^2}

    h\prime(x)=\frac{4(2x^2-4x^6)^3(4x-24x^5)-(4x)(2x^2-4x^6)^4}{(2+2x^2)}


    I am 75% sure I think I got this one correct as well

    f(x)=ln(2^x-3^{\frac{1}{2}}
    f\prime(x)=\frac{(ln2(2^x)-\frac{3}{2\sqrt{x}}}{(2^x-3\sqrt{x})}

    This also

    y=\sqrt{3e^x-x^\pi}

    y\prime=\frac{1}{2}(3e^x-x^\pi) * (3e^x-\pi^{\pi-1})

    Sorry
    Last edited by sk8erboyla2004; March 14th 2009 at 08:50 PM.
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    Quote Originally Posted by sk8erboyla2004 View Post
    Darn I need mad practice, ugh.

    Thanks also check this one it has far as i got and im unsure of how to future simply it bc i maybe be wrong

    h(x)=\frac{(2x^2-4x^6)^4}{2+2x^2}

    h\prime(x)=\frac{4(2x^2-4x^6)^3(4x-24x^5)(2+2x^2)-(4x)(2x^2-4x^6)^4}{(2+2x^2)^2}

    h\prime(x)=\frac{4(2x^2-4x^6)^3(4x-24x^5)-(4x)(2x^2-4x^6)^4}{(2+2x^2)}
    Like I say, it's always easier to break it into smaller parts.

    h = \frac{(2x^2 - 4x^6)^4}{2 + 2x^2}

    This is a quotient, so start with the quotient rule

    \frac{dh}{dx} = \frac{(2 + 2x)\frac{d}{dx}[(2x^2 - 4x^6)^4] - (2x^2 - 4x^6)^4\frac{d}{dx}(2 + 2x^2)}{(2 + 2x^2)^2}

    For the difficult looking derivative, use the chain rule.

    y = (2x^2 - 4x^6)^4

    Let u = 2x^2 - 4x^6 so that y = u^4

    \frac{du}{dx} = 4x - 24x^5, \frac{dy}{du} = 4u^3 = 4(2x^2 - 4x^6)^3.

    So \frac{dy}{dx} = 4(4x - 24x^5)(2x^2 - 4x^6)^3.


    Therefore

    \frac{dh}{dx} = \frac{4(2 + 2x)(4x - 24x^5)(2x^2 - 4x^6)^3 - 4x(2x^2- 4x^6)^4}{(2 + 2x)^2}.
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    I had to make a separate post for this one

    y=(\frac{x^3+2x}{e^{x^2}})^3

    y\prime=3(\frac{x^3+2x}{e^{x^2}})[(3x^2+2)(e^{x^2}-(x^3+2)(e^{x^2})(2x)]

    if this is right how can i further simplify it?
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    Quote Originally Posted by Prove It View Post
    Like I say, it's always easier to break it into smaller parts.

    h = \frac{(2x^2 - 4x^6)^4}{2 + 2x^2}

    This is a quotient, so start with the quotient rule

    \frac{dh}{dx} = \frac{(2 + 2x)\frac{d}{dx}[(2x^2 - 4x^6)^4] - (2x^2 - 4x^6)^4\frac{d}{dx}(2 + 2x^2)}{(2 + 2x^2)^2}

    For the difficult looking derivative, use the chain rule.

    y = (2x^2 - 4x^6)^4

    Let u = 2x^2 - 4x^6 so that y = u^4

    \frac{du}{dx} = 4x - 24x^5, \frac{dy}{du} = 4u^3 = 4(2x^2 - 4x^6)^3.

    So \frac{dy}{dx} = 4(4x - 24x^5)(2x^2 - 4x^6)^3.


    Therefore

    \frac{dh}{dx} = \frac{4(2 + 2x)(4x - 24x^5)(2x^2 - 4x^6)^3 - 4x(2x^2- 4x^6)^4}{(2 + 2x)^2}.
    SO I was somewhere close?
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    Quote Originally Posted by sk8erboyla2004 View Post
    I had to make a separate post for this one

    y=(\frac{x^3+2x}{e^{x^2}})^3

    y\prime=3(\frac{x^3+2x}{e^{x^2}})[(3x^2+2)(e^{x^2}-(x^3+2)(e^{x^2})(2x)]

    if this is right how can i further simplify it?
    The easiest way to make mistakes is to skip steps.

    Write down EVERYTHING.

    y = \left(\frac{x^3 + 2x}{e^{x^2}}\right)^3.

    You should be able to see that it is a quotient taken to a power. So use the chain rule first (as the "final" thing done was to take it to a power...)

    So Let u = \frac{x^3 + 2x}{e^{x^2}} so that y = u^3.

    \frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du}.


    \frac{dy}{du} is easy, it's

    3u^2 = 3\left(\frac{x^3 + 2x}{e^{x^2}}\right)^2.


    \frac{du}{dx} is harder, as it's a quotient.

    \frac{du}{dx} = \frac{e^{x^2}\frac{d}{dx}(x^3 + 2x) - (x^3 + 2x)\frac{d}{dx}(e^{x^2})}{(e^{x^2})^2}

     = \frac{e^{x^2}(3x^2 + 2) - 2xe^{x^2}(x^3 + 2x)}{e^{2x^2}}.


    Therefore

    \frac{dy}{dx} = \frac{3[e^{x^2}(3x^2 + 2) - 2xe^{x^2}(x^3 + 2x)]}{e^{2x^2}}\left(\frac{x^3 + 2x}{e^{x^2}}\right)^2.
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    I really appreciate this bro, with each problem Im getting better with the last one you just posted i forgot to add the /g(x)^2 to the bottom

    Thanks Ill have many more to come lol!

    like this one

    r(p)=tan^3(4p)

    r\prime(p)=\frac{4}{cos^3(4p)}
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