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    Differentiate 6sin\sqrt x
    I did this:
    u= x^{1/2} and y=6sinu

    \frac{dy}{dx}=6cosu \frac{du}{dx}=1/2x^{-1/2}

    The answer is meant to be 3cossqrtx / sqrtx
    How do you get that? If you times it together it doesnt come to that ? :|

    P.S Merry Christmas everyone (and for those not celebrating) Merry Saturday!
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    (Original post by Nkhan)
    Differentiate 6sin\sqrt x
    I did this:
    u= x^{1/2} and y=6sinu

    \frac{dy}{dx}=6cosu \frac{du}{dx}=1/2x^{-1/2}

    The answer is meant to be 3cossqrtx / sqrtx
    How do you get that? If you times it together it doesnt come to that ? :|

    P.S Merry Christmas everyone (and for those not celebrating) Merry Saturday!
    (\frac{1}{2}x^{-\frac{1}{2}})(6cos(x^\frac{1}{2}  ))

    That's the same as

    (\frac{1}{2})(6)(x^{-\frac{1}{2}})(cos(x^\frac{1}{2})  )

    Because multiplication is multiplication - it doesn't matter if you do 6x8 or 8x6. So it doesn't matter if you do (1/2)*(x^-1/2)*(6) or (x^-1/2)*(6)*(1/2) or (1/2)*(6)*(x^-1/2).

    x^{-2} = \frac{1}{x^2}

    x^{-\frac{5}{6}}=\frac{1}{x^{\frac{5  }{6}}}

    You should probably be able to figure out what x^{-\frac{1}{2}} is as a fraction now if you already didn't know
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    (Original post by Chelle-belle)
    (\frac{1}{2}x^{-\frac{1}{2}})(6cos(x^\frac{1}{2}  ))

    That's the same as

    (\frac{1}{2})(6)(x^{-\frac{1}{2}})(cos(x^\frac{1}{2})  )

    Because multiplication is multiplication - it doesn't matter if you do 6x8 or 8x6. So it doesn't matter if you do (1/2)*(x^-1/2)*(6) or (x^-1/2)*(6)*(1/2) or (1/2)*(6)*(x^-1/2).

    x^{-2} = \frac{1}{x^2}

    x^{-\frac{5}{6}}=\frac{1}{x^{\frac{5  }{6}}}

    You should probably be able to figure out what x^{-\frac{1}{2}} is as a fraction now if you already didn't know
    Thankyou I get it now
 
 
 
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