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Differentiating inverse trigonometric function Watch

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    f( x)=arccos^2x


    Let u=x and v=arccosx^2



    \frac{d}{dx} uv=v \frac{du}{dx} +u \frac{dv}{dx}



    =arccosx^2+x(\frac{d}{dx} (arccosx^2))



    I already worked out that,



    \frac{d}{dx} arccosx^2=-\frac{2x}{\sqrt{1-x^4}}



    So,


    \frac{d}{dx} uv=arccosx^2+x(-\frac{2x}{\sqrt{1-x^4}} )



    =arccosx^2-\frac{2x^2}{\sqrt{1-x^4}}




    Where have I gone wrong, as I know my answer is supposed to be,




    arccosx^2-\frac{2xarccosx}{\sqrt{1-x^4}}



    Thanks.
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    (Original post by hollywoodbudgie)
    .
    arccos^2(x) is just one function. think of it as (arccosx)^2

    so it's the chain rule, not the product rule
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    Do you mean f(x)=cos^{-1}(x^2) or f(x)=(cos^{-1}(x))^2?

    Hopefully the former, in which case let y=f(x) for ease of notation, then x^2=cos(y) and differentiate implictly. Similarly for the latter, but it's a bit uglier
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    (Original post by hollywoodbudgie)
    Thanks.
    No that's right.

     \dfrac{d}{dx} \left(xarccos(x^2)\right) = arccos(x^2)-\dfrac{2x^2}{\sqrt{1-x^4}}

     \dfrac{d}{dx} \left(xarccos^2(x)\right) = arccos^2(x)-\dfrac{2xarccosx}{\sqrt{1-x^2}}
 
 
 
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