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Math Help - Confunsing Chain Rules and more

  1. #16
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    Quote Originally Posted by sk8erboyla2004 View Post
    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)}
    You keep making mistakes because you're not doing all the steps.

    Here r = \tan^3{(4p)} = [\tan{(4p)}]^3.

    It's a function of a function, so use the chain rule.

    Let u = \tan{(4p)} so that r = u^3.


    \frac{du}{dp} = \frac{4}{\cos^2{(4p)}}

    \frac{dr}{du} = 3u^2 = 3\tan^2{(4p)}.

    So \frac{dr}{dp} = \frac{12\tan^2{(4p)}}{\cos^2{(4p)}}.
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  2. #17
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    h(t)=\arcsin(t^3)-ln(t^2+1)

    h\prime(t)=\frac{3t^2}{\sqrt{1-t^6}}-\frac{2t}{t^2+1}

    ?


    y=(csc(x))^20

    y\prime=20(csc(x))^{19}*-csc(x)cot(x)
    Last edited by sk8erboyla2004; March 15th 2009 at 01:36 PM.
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  3. #18
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    Quote Originally Posted by sk8erboyla2004 View Post
    Thanks

    h(t)=\arcsin(t^3)-ln(t^2+1)

    h\prime(t)=\frac{3t^2}{\sqrt{1-t^6}}-\frac{2t}{t^2+1}

    ?


    y=(csc(x))^20

    y\prime=20(csc(x))^{19}*-csc(x)cot(x)
    The first one looks ok...

    For the second, remember \csc{x} = \frac{1}{\sin{x}} = (\sin{x})^{-1}.

    So y = (\csc{x})^{20} = (\sin{x})^{-20}

    Let u = \sin{x} so that y = u^{-20}.

    \frac{du}{dx} = \cos{x}, \frac{dy}{du} = -20u^{-21} = -20(\sin{x})^{-21}.

    So \frac{dy}{dx} = -20\cos{x}\sin^{-21}{x} = -20\cos{x}\csc^{21}{x}.
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  4. #19
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    Thanks mate! You really know your stuff the more you help the more I learn.

    If you dont mind Im gonna keep this coming at for a good bit?


    y=-sin(9x^3+21)

    y\prime=cos(9x^3+21)*(27x^2)

    ? It seems im getting a tad better I think I have the chain rule down now I have to venture into derivatives of trig and inverse trig functions for this assignment

    y=sin(sin(x))

    [tex]y\prime=cos(sin(x))*cos(x)[/MATH


    y=cos(sec(x))
    y\prime=-sin(sec(x))*sec(x)tan(x)

    Im sure you have any other what but is this way correct as well?
    Last edited by sk8erboyla2004; March 15th 2009 at 03:04 PM.
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  5. #20
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    Quote Originally Posted by sk8erboyla2004 View Post
    Thanks mate! You really know your stuff the more you help the more I learn.

    If you dont mind Im gonna keep this coming at for a good bit?


    y=-sin(9x^3+21)

    y\prime=cos(9x^3+21)*(27x^2)

    ? It seems im getting a tad better I think I have the chain rule down now I have to venture into derivatives of trig and inverse trig functions for this assignment

    y=sin(sin(x))

    [tex]y\prime=cos(sin(x))*cos(x)[/MATH


    y=cos(sec(x))
    y\prime=-sin(sec(x))*sec(x)tan(x)

    Im sure you have any other what but is this way correct as well?
    The first two are correct.

    y = \cos{\sec{x}} = \cos{\frac{1}{\cos{x}}} = \cos{[(\cos{x})^{-1}]}.

    Here you have a function of a function of a function, so you're going to have to use the chain rule twice.

    Let u = (\cos{x})^{-1} so that y = \cos{u}

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


    \frac{dy}{du} = -\sin{u} = -\sin{[(\cos{x})^{-1}]}. That part was easy.

    \frac{du}{dx} is not quite so easy.

    Let v = \cos{x} so that u = v^{-1}.

    \frac{dv}{dx} = -\sin{x}

    \frac{du}{dv} = -v^{-2} = -(\cos{x})^{-2}.

    So \frac{du}{dx} = \sin{x}(\cos{x})^{-2}.


    Therefore

    \frac{dy}{dx} = -\sin{x}(\cos{x})^{-2}\sin{[(\cos{x})^{-1}]}=-\tan{x}\sec{x}\sin{(\sec{x})}.

    So that's right as well
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  6. #21
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    Finally, Thanks man you were an amazing help!

    Count this one solved for now. I just have good bit of problems left I wish I could get them checked but there are too many a las oh well, i feel confident now

    Thanks Prove it reps and thanks coming your way!

    Edit: Nvm I need a bit more checking lol

    y=e^{15csc(5x)}
    y\prime=15csc(5)e^{15csc(5x)}*75csc(5x)cot(5x)
    Last edited by sk8erboyla2004; March 15th 2009 at 04:16 PM.
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  7. #22
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    Quote Originally Posted by sk8erboyla2004 View Post
    Finally, Thanks man you were an amazing help!

    Count this one solved for now. I just have good bit of problems left I wish I could get them checked but there are too many a las oh well, i feel confident now

    Thanks Prove it reps and thanks coming your way!

    Edit: Nvm I need a bit more checking lol

    y=e^{15csc(5x)}
    y\prime=15csc(5)e^{15csc(5x)}*75csc(5x)cot(5x)
    y = e^{15\csc{(5x)}} = e^{15(\sin{(5x)})^{-1}}.

    Let u = 15(\sin{(5x)})^{-1} so that y = e^u.

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

    \frac{dy}{du} = e^u = e^{15(\sin{(5x)})^{-1}}


    Let v = \sin{(5x)} so that u = 15v^{-1}.

    \frac{dv}{dx} = 5\cos{(5x)}

    \frac{du}{dv} = -15v^{-2} = -15(\sin{(5x)})^{-2}

    So \frac{du}{dx} = -75\cos{(5x)}(\sin{(5x)})^{-2}.

    What does \frac{dy}{dx} equal?
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  8. #23
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    e^{15csc(5x)}*-75csc(5x)cot(5x)

    ?
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  9. #24
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    Quote Originally Posted by sk8erboyla2004 View Post
    e^{15csc(5x)}*-75csc(5x)cot(5x)

    ?
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