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Math Help - chain rule differentiation check!

  1. #1
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    chain rule differentiation check!

    derivative of:

    <br /> <br />
y = sin(tan(sinx))<br />

    chain rule:

    <br /> <br />
y' = cos(cosxsec^2x(sinx))<br />

    i got upto there but im ment to differntiate this but im not sure what to do inside that bracket to differentiate that.
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  2. #2
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    Krizalid's Avatar
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    y'=\cos (\tan (\sin x))\cdot \left( \tan (\sin x) \right)', and chain rule again.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    y'=\cos (\tan (\sin x))\cdot \left( \tan (\sin x) \right)', and chain rule again.
    hrmm not to sure what u mean. first i take the derivative of sin(tan(sinx))
    which is:

    <br /> <br />
cos(tan(sinx)) \cdot sec^2(sinx)<br /> <br />

    then derivative of tan(sinx)

    <br /> <br />
sec^2(sinx) \cdot cosx<br /> <br />

    so answer:

    <br /> <br />
cos(tan(sinx)) \cdot sec^2(sinx) \cdot sec^2(sinx) \cdot cosx<br /> <br />
    this correcT?
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  4. #4
    Super Member 11rdc11's Avatar
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    Nope but almost

    cos(tan(sinx)) \cdot sec^2(sinx) \cdot cosx
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  5. #5
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    Quote Originally Posted by 11rdc11 View Post
    Nope but almost

    cos(tan(sinx)) \cdot sec^2(sinx) \cdot cosx
    thank you
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