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Math Help - Calculus Help Please!

  1. #1
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    Exclamation Calculus Help Please!

    I'm studying for a test, and I don't understand how to do this problem. Could anyone help me?

    Find the derivative dy/dx if y equals the integral from sin(x) to 5 of 1/u^3 du. Be sure to show your work and give the statement of any theorems that you use.
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  2. #2
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    I can tell you the answer: -\cot x \csc^2 x.
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  3. #3
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    Quote Originally Posted by juicysharpie View Post
    I'm studying for a test, and I don't understand how to do this problem. Could anyone help me?

    Find the derivative dy/dx if y equals the integral from sin(x) to 5 of 1/u^3 du. Be sure to show your work and give the statement of any theorems that you use.
    y = \int_{\sin x}^5 \frac{1}{u^3} \, du = - \int^{\sin x}_5 \frac{1}{u^3} \, du.


    Let t = \sin x and use the chain rule:


    \frac{dy}{dx} = \frac{d}{dt} \left[- \int^{t}_5 \frac{1}{u^3} \, du \right] \cdot \frac{dt}{dx}


    = -\frac{1}{t^3} \cdot \cos x = -\frac{1}{\sin^3 x} \cdot \cos x = \frac{\cos x}{\sin^3 x}.


    Your job is to identify and name the theorems that I used.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    = -\frac{1}{t^3} \cdot \cos x = -\frac{1}{\sin^3 x} \cdot \cos x = \frac{\cos x}{\sin^3 x}.
    Looks good except you dropped a minus sign on the last step.
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  5. #5
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    Question

    Quote Originally Posted by mr fantastic View Post
    \frac{dy}{dx} = \frac{d}{dt} \left[- \int^{t}_5 \frac{1}{u^3} \, du \right] \cdot \frac{dt}{dx}
    I'm sorry, could you please explain how you got to this step? I don't quite understand how you did it.

    Thanks for helping!!
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  6. #6
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    Quote Originally Posted by juicysharpie View Post
    I'm sorry, could you please explain how you got to this step? I don't quite understand how you did it.

    Thanks for helping!!
    Review chain rule in your class notes or textbook.
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