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Math Help - Sin/Cos Differentiation

  1. #1
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    Sin/Cos Differentiation

    Here are a couple of sin cos differentiation questions. My teacher has gone over the rules for a bit but hasn't shown how to do the actual equation questions.

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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Here are a couple of sin cos differentiation questions. My teacher has gone over the rules for a bit but hasn't shown how to do the actual equation questions.

    f(x) = cos(pi/x)

    put g(x)=pi/x, then:

    f(x) = cos(g(x))

    Now we use the chain rule:

    f'(x) = -sin(g(x)) g'(x) = -sin(pi/x) (-pi/x^2)

    so f'(2) = sin(pi/2) (pi/4) ~= 0.785

    RonL
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Here are a couple of sin cos differentiation questions. My teacher has gone over the rules for a bit but hasn't shown how to do the actual equation questions.

    y = x cos(x)/(1+e^x)

    we find dy/dx using the product rule and the quotient rule:

    dy/dx = d/dx[x/(1+e^x)] cos(x) + [x/(1+e^x)] d/dx cos(x)

    Now you need to use the quotient rule to do the first of the derivatives on
    the right above. Me I don't know the quotient rule so I will use the product
    rule a second time:

    d/dx[x/(1+e^x)] = 1/(1+e^x) - x e^x /(1+e^x)^2 = (1 - e^x (x - 1))/(e^x + 1)^2

    so:

    dy/dx = [(1 - e^x (x - 1))/(e^x + 1)^2] cos(x) - [x/(1+e^x)] sin(x)

    RonL
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    How do you differentiate this one since it has both sine and cosine in it?

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    Quote Originally Posted by SportfreundeKeaneKent View Post
    How do you differentiate this one since it has both sine and cosine in it?

    Individual terms and the chain rule:
    d/dx[sin^2(x)] = 2*sin(x)*cos(x)
    d/dx[cos(2x - 1)] = -sin(2x - 1) * 2

    So
    dy/dx = 12*sin(x)*cos(x) - 2sin(2x - 1)

    -Dan
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