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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)
IsTim now you're here for math too? giggle, shiny and math?
y=x^n dy/dx = nx^(n-1) this is the property I use
Truly, I don't really understand what derivatives is. @Ishaan94 Pardon?
The Power Rule. Ok.
It's ok. I don't understand (your speech and derivatives; either/or) though.
If you can, please describe each step thoroughly. Thank you.
What's happening at dy/dx = 5(3/2)(x^(3/2 -1)) ?
Or, how did it come to that?
y=ax^n dy/dx =a( nx^(n-1)) put a=1 and n=3/2
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.
@CoCoTsoi When you have time, of course.
That was just showing you an of how to work the power rule. I rewrote the sqrt as x^1/2
|dw:1335434722081:dw| That was the answer in the textbook. How do I convert to that?
A-ha, that wasn't 5 times sqrt X (etc...) It was the fifth root
|dw:1335424089162:dw| was the original question