# Finding derivative of a function

• Jul 21st 2012, 12:31 AM
astuart
Finding derivative of a function
Hello,

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

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

Using the form $\displaystyle (d/dx)f(x)$, 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

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

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
• Jul 21st 2012, 12:42 AM
earboth
Re: Finding derivative of a function
Quote:

Originally Posted by astuart
Hello,

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

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

Using the form $\displaystyle (d/dx)f(x)$, 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

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

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:

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

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

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

To evaluate the volume or the surface area of a sphere with $\displaystyle r = \frac23$ you only have to plug in this value into these equations
• Jul 21st 2012, 02:09 AM
astuart
Re: Finding derivative of a function
Quote:

Originally Posted by earboth
What you posted is the term to calculate the volume of a sphere:

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

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

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

To evaluate the volume or the surface area of a sphere with $\displaystyle r = \frac23$ 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 $\displaystyle (16(pi)/9)cm^3/cm$ - I'm just not sure how to get that answer...
• Jul 21st 2012, 04:35 AM
skeeter
Re: Finding derivative of a function
Quote:

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

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

$\displaystyle V'\left(\frac{2}{3}\right) = 4 \pi \left(\frac{2}{3}\right)^2 = \, ?$
• Jul 22nd 2012, 01:11 AM
astuart
Re: Finding derivative of a function
Quote:

Originally Posted by skeeter
$\displaystyle V'(r) = 4\pi r^2$

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

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.

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

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

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

Thanks again, the Pi threw me off..