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Mathematics 8 Online
OpenStudy (anonymous):

Find the derivative of 5pi/3

hartnn (hartnn):

@Floraaa Hi :) \(\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\) since 5pi/3 is constant , its derivative =0 are you sure the question states, 5pi/3 ?

hartnn (hartnn):

or its something like sin 5pi/3 or cos 5pi/3 ??

OpenStudy (anonymous):

It's 0 5π^3 is a constant, most people use π to represent the approximation to 3.14159... So what would you have said the derivative is of 5(3.14159...)^3 ? Or what would you say is the derivative of y = 135

OpenStudy (anonymous):

got it?

OpenStudy (anonymous):

Oh okay i get it. i just have to make the whole equation equal to zero to find the complete derivative. Thanks(:

OpenStudy (anonymous):

ok

hartnn (hartnn):

umm...not really, can you post the complete question ?

OpenStudy (anonymous):

I'll attach it.

OpenStudy (anonymous):

Here it is.

hartnn (hartnn):

you need to find the derivative of 'r' with respect to 'h' (and not the derivative of 5pi/3) were u able to find it ?

OpenStudy (anonymous):

I'm working on it.

hartnn (hartnn):

i'll wait, if you have doubts, u can ask :)

OpenStudy (anonymous):

I'm not getting the right answer. I think i'm doing it wrong.

hartnn (hartnn):

can u show me what u did ?

hartnn (hartnn):

i'll spot the error....this way it'll be faster..

OpenStudy (anonymous):

I cant take a picture of my work. For the top we will be using the product rule and the implicit differentiation thing right?

hartnn (hartnn):

yes, and you put V =5pi/3, right ? (note the pi's getting cancelled...)

hartnn (hartnn):

did u get \(5=h^2(3r-h)\)

hartnn (hartnn):

then instead of product rule first, you can do this : 5 = 3rh^2-h^3 can you differentiate this ?

OpenStudy (anonymous):

Yes I think I got it. Thanks(:

hartnn (hartnn):

welcome ^_^

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