Quick differentiation help please.... Watch

Nkhan
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Report Thread starter 8 years ago
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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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Chelle-belle
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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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Nkhan
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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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