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Math Help - Chain rule stuff

  1. #1
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    Chain rule stuff

    Fine the derivative of the function:
    1. g(x)=(1+4x)^5 (3+x-x^2)^8
    2. y(x)=2^(sin(pie)x)
    3. (1+cos^2x)^6

    i dont get this **** and my test is on thursday.
    Last edited by ThePerfectHacker; October 17th 2006 at 04:49 PM.
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  2. #2
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    Quote Originally Posted by jeph View Post
    Fine the derivative of the function:
    1. g(x)=(1+4x)^5 (3+x-x^2)^8
    First use the product rule,
    [(1+4x)^5]'(3+x-x^2)+(1+4x)^5[(3+x-x^2)^8]'
    Now use the chain rule,
    [5*4(1+4x)^4](3+x-x^2)+(1+4x)^5[8(1-2x)(3+x-x^2)^7]
    Simplify,
    20(1+4x)^4(3+x-x^2)+8(1+4x)^5(1-2x)(3+x-x^2)^7
    Factor,
    (1+4x)^4(3+x-x^2)[20+8(1+4x)(1-2x)(3+x-x^2)^6]
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  3. #3
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    Quote Originally Posted by jeph View Post
    2. y(x)=2^(sin(pie)x)
    We can express 2^x as e^(ln 2 * x)
    Thus,
    2^[sin (pi x)]=e^[ln 2 *sin (pi x)]
    Now, the chain rule says that,
    e^{f(x)} then the derivative is, f'(x)e^{f(x)}
    That is take derivate of exponent and bring it down.
    In this case the exponent is,
    ln 2* sin(pi x) where ln 2 is a constant function,
    The derivative is after using chain rule,
    pi*ln 2*cos(pi x)
    Thus, the answer is,
    (pi*ln 2*cos(pi x))e^[ ln 2*sin (pi x)]
    But, the 'e' exponent simplifies to (look at beginning of post),
    (pi*ln 2*cos(pi x))2^[sin (pi x)]
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    Quote Originally Posted by jeph View Post
    3. (1+cos^2x)^6

    .
    You use a double chain rule here.
    The inner function is,
    1+[cos x]^2
    The derivative of this function is,
    -2 sin x cos x=-sin 2x (double angle)
    Thus, derivative of
    (1+cos^2 x)^6 is,
    Inner function derivative which we found to be,
    -sin 2x
    Times outer function derivative,
    6(1+cos^2 x)^5
    Multiply them out,
    -6*sin 2x(1+cos^2 x)^5
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  5. #5
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    sorry about using that word, i didnt think it would have been considered inappropriate =/

    can you show me how
    1+[cos x]^2 = -2 sin x cos x

    this whole time i tohught i wasnt getting the chain rule, but this is the part that got me.

    the answer given was -12cosxsinx(1+cos^2x)^5
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  6. #6
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    Quote Originally Posted by jeph View Post
    sorry about using that word, i didnt think it would have been considered inappropriate =/

    can you show me how
    1+[cos x]^2 = -2 sin x cos x

    this whole time i tohught i wasnt getting the chain rule, but this is the part that got me.

    the answer given was -12cosxsinx(1+cos^2x)^5
    New questions should be posted in new threads!!

    Are you asking how to take the derivative of 1 + [cos(x)]^2? Because

    1+[cos x]^2 is most certainly NOT -2 sin x cos x. (For starters the LHS is always positive.)

    d/dx{1 + [cos(x)]^2} = 0 + d/dx{[cos(x)]^2}

    Your "outer function" is y^2 where y = cos(x) and your "inner function" is cos(x), so take the derivative of y^2 (2*y) and multiply that by the derivative of cos(x): ( [f(g(x))]' = f'(g(x))*g'(x) )
    d/dx{[cos(x)]^2} = 2*cos(x)* -sin(x) = -2sin(x)cos(x)

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    New questions should be posted in new threads!!
    I do not consider that to be a violation. Since that question is asking in my solution which the user did not understand.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I do not consider that to be a violation. Since that question is asking in my solution which the user did not understand.
    (Shakes his head in despair) Caught not reading the whole question again wasn't I? I tell you, I'm on a roll.

    -Dan
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