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Math Help - Help with Derivative

  1. #1
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    Help with Derivative

    I am having a few problems with two derivative problems.

    The first problem is to find the derivative of sqr(1-25x^2)arcos(5x) and simplify if possible. I havent got the right answer after many tries i assumed you needed to use the product rule to compute this derivative.

    The next problem is to find the tangent line at the point (-3sqr(3),1) for the curve x^(2/3) + y^(2/3) = 4.

    If anyone could show me how to do this it would be much appreciated.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by NickytheNine View Post
    I am having a few problems with two derivative problems.

    The first problem is to find the derivative of sqr(1-25x^2)arcos(5x) and simplify if possible. I havent got the right answer after many tries i assumed you needed to use the product rule to compute this derivative.
    You will need product rule for this....

    Note that \frac{\,d}{\,dx}\left[\cos^{-1}u\right]=-\frac{\,d}{\,dx}\left[\sin^{-1}u\right]=-\frac{1}{\sqrt{1-u^2}}\cdot\frac{\,du}{\,dx}

    So when you apply product rule, you should end up with -\frac{25}{\sqrt{1-25x^2}}\cos^{-1}(5x)+\sqrt{1-25x^2}\cdot\left[-\frac{5}{\sqrt{1-25x^2}}\right]

    Do you see why this is the case? I leave it for you to simplify this.

    The next problem is to find the tangent line at the point (-3sqr(3),1) for the curve x^(2/3) + y^(2/3) = 4.

    If anyone could show me how to do this it would be much appreciated.
    For this one, use implicit differentiation...

    You should end up with \frac{\,dy}{\,dx}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}} [I leave it for you to show that this is the case].

    Then plug your point in to find the slope of the tangent line. Then plug in all the known values into the point-slope equation for a line and get it into y=mx+b form.

    Can you try this one and see what you get?

    --Chris
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