Finding derivative of a function

**astuart** · July 21st 2012, 12:31 AM

Hello,

I'm supposed to find the derivative of the following.

$(4/3)Pi (r)^3$

Using the form $(d/dx)f(x)$ , or turning F(x) into F'(x) (Prime).

Anyway, I'm supposed to solve r=2/3 but I'm not sure how I'm supposed to turn this function into F'(x).

I think I'm supposed to do something like

$(4/3)Pi (3r)^2$

But I'm not sure how I should be applying the exponent when r = 2/3 - or whether I'm on the right track in that case.

Any tips?

Cheers

**earboth** · July 21st 2012, 12:42 AM

Originally Posted by astuart

Hello,

I'm supposed to find the derivative of the following.

$(4/3)Pi (r)^3$

Using the form $(d/dx)f(x)$ , or turning F(x) into F'(x) (Prime).

Anyway, I'm supposed to solve(?) r=2/3 <-- this equation is already solved
but I'm not sure how I'm supposed to turn this function into F'(x).

I think I'm supposed to do something like

$(4/3)Pi (3r)^2$

But I'm not sure how I should be applying the exponent when r = 2/3 - or whether I'm on the right track in that case.

Any tips?

Cheers

What you posted is the term to calculate the volume of a sphere:

$V(r)=\frac43 \pi r^3$

Then $V'(r)=\frac{dV}{dr}= \frac43 \pi r^2 \cdot 3=4 \pi r^2=A_s(r)$

which is the equation to determine the surface area of a sphere.

To evaluate the volume or the surface area of a sphere with $r = \frac23$ you only have to plug in this value into these equations

**astuart** · July 21st 2012, 02:09 AM

Originally Posted by earboth

What you posted is the term to calculate the volume of a sphere:

$V(r)=\frac43 \pi r^3$

Then $V'(r)=\frac{dV}{dr}= \frac43 \pi r^2 \cdot 3=4 \pi r^2=A_s(r)$

which is the equation to determine the surface area of a sphere.

To evaluate the volume or the surface area of a sphere with $r = \frac23$ you only have to plug in this value into these equations

I knew it was the volume of a sphere but I just wasn't sure how to put that into V'.

Apparently the answer is $(16(pi)/9)cm^3/cm$ - I'm just not sure how to get that answer...

**skeeter** · July 21st 2012, 04:35 AM

Originally Posted by astuart

I knew it was the volume of a sphere but I just wasn't sure how to put that into V'.

Apparently the answer is $(16(pi)/9)cm^3/cm$ - I'm just not sure how to get that answer...

$V'(r) = 4\pi r^2$

$V'\left(\frac{2}{3}\right) = 4 \pi \left(\frac{2}{3}\right)^2 = \, ?$

**astuart** · July 22nd 2012, 01:11 AM

Originally Posted by skeeter

$V'(r) = 4\pi r^2$

$V'\left(\frac{2}{3}\right) = 4 \pi \left(\frac{2}{3}\right)^2 = \, ?$

Ah, great, thanks for the help guys! I wasn't sure whether the Pi was considered a constant, and therefore can be excluded from the V' function, or whether a different rule applies for 'special' constants such as Pi.

$4/3 * 3/1 = 4$ to get that section, then -1 from the exponent.

$(2/3)^2 = 4/9$

$4/9 * 4/1 = 16(Pi)/9$

Thanks again, the Pi threw me off..

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