IsTim
Determine dy/dx for y=5sqrt(x^3)



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CoCoTsoi
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y=5sqrt(x^3)
= 5(x^(3/2))
dy/dx = 5(3/2)(x^(3/2 1))
= 15/2 (x^1/2)

Ishaan94
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IsTim now you're here for math too? giggle, shiny and math?

CoCoTsoi
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y=x^n
dy/dx = nx^(n1)
this is the property I use

IsTim
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Truly, I don't really understand what derivatives is.
@Ishaan94 Pardon?

IsTim
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The Power Rule. Ok.

Ishaan94
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Sorry.

IsTim
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It's ok. I don't understand (your speech and derivatives; either/or) though.

IsTim
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If you can, please describe each step thoroughly. Thank you.

IsTim
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OH well.

IsTim
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What's happening at dy/dx = 5(3/2)(x^(3/2 1)) ?

IsTim
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Or, how did it come to that?

CoCoTsoi
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y=ax^n
dy/dx =a( nx^(n1))
put a=1 and n=3/2

IsTim
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Uhm, sorry to ask, but could you type dy/dx = 5(3/2)(x^(3/2 1)) in Latex (The Equation Editor)? I don't think I'm understanding it correctly.

IsTim
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@CoCoTsoi When you have time, of course.

CoCoTsoi
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dw:1335423732321:dw

IsTim
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Thanks.

Maekin
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dw:1335423870842:dwdw:1335423900291:dw

Maekin
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That was just showing you an of how to work the power rule. I rewrote the sqrt as x^1/2

IsTim
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dw:1335434722081:dw
That was the answer in the textbook. How do I convert to that?

Maekin
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Aha, that wasn't 5 times sqrt X (etc...) It was the fifth root

Maekin
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dw:1335424089162:dw was the original question

IsTim
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Yes.