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Thread: Trigonometric Derivative

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    Trigonometric Derivative

    What is the derivative of $\displaystyle sin^\frac {1}{3} x $?
    Actually part of a bigger problem, but I think I can tackle it if I just know this.
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    Quote Originally Posted by seuzy13 View Post
    What is the derivative of $\displaystyle sin^\frac {1}{3} x $?
    Actually part of a bigger problem, but I think I can tackle it if I just know this.
    let $\displaystyle t=y^{\frac{1}{3}}$ and $\displaystyle y=\sin x $ we need dt/dx

    $\displaystyle dt/dx = dt/dy \cdot dy/dx $

    $\displaystyle dt/dx = 1/3y^{\frac{-2}{3}} \cdot \cos x $ sub y value

    $\displaystyle dt/dx = 1/3(\sin x)^{\frac{-2}{3}}(\cos x )$

    $\displaystyle dt/dx = \frac{\cos x }{3(\sin ^{\frac{2}{3}})}$

    that's it
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