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Math Help - Equation of the tangent

  1. #1
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    Smile Equation of the tangent

    Determine the equation of the tangent to the curve f(x)=x*sqrt(1-x^2) at x=1/2 ???

    What I have done:
    y=sqrt3/4
    ((1/2),(sqrt3/4))

    I am having problem finding the derivative f'(1/2). I know that f'(1/2) = 1/sqrt3

    Can anyone help me by showing me how they have done f'(1/2) step by step
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  2. #2
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    Quote Originally Posted by sf1903 View Post
    Determine the equation of the tangent to the curve f(x)=x*sqrt(1-x^2) at x=1/2 ???

    What I have done:

    y=\frac{\sqrt{3}}{4}

    (x,y)=\left(\frac{1}{2},\frac{\sqrt{3}}{4}\right)

    I am having problem finding the derivative f'(1/2). I know that f'(1/2) = 1/sqrt3

    Can anyone help me by showing me how they have done f'(1/2) step by step
    Use the product rule of differentiation..


    \displaystyle\ f(x)=x\sqrt{1-x^2}\Rightarrow\ \frac{dy}{dx}=x\frac{d}{dx}\sqrt{1-x^2}+\sqrt{1-x^2}\ \frac{dx}{dx}

    Use the chain rule to differentiate \sqrt{1-x^2}


    \displaystyle\Rightarrow\ \frac{dy}{dx}=\frac{x}{2\sqrt{1-x^2}}(-2x)+\sqrt{1-x^2}

    Now substitute x=0.5 to get

    \displaystyle\ f'(0.5)=\frac{1}{\sqrt{3}}
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  3. #3
    MHF Contributor

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    It might help you to write f(x) as f(x)= x(1- x^2)^{1/2}. You will need to use the power rule, the product rule, and the chain rule.
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