I can't differentiate this

[Air conditionner]

Hi,
I'm working on calculus problems, and there is one equation I can't differentiate. It's this one:

Because I've done it using the power and the chain rule, I got the following:

But the app I use to verify the answers constantly gives this one and only answer:

I'm completely lost.
What is the derivative of the function? And do you know why the computer answers v' = 0 ?
Thank you,
Air Conditioner

 

[pka]

Lets play the back-of-the-book game. Look up the answer to this and see:
[imath]D_r\left(2\pi r^2\sqrt{R^2-r^2}\right)=\dfrac{4\pi r R^2-6\pi r^3}{\sqrt{R^2-r^2}}[/imath] SEE THIS
Now your job is to find out why is that the answer?

 

[Air conditionner]

I really tried to understand the equation you wrote, but I am lost. Could you please explain further? I understand where the left side comes from, but I can't get to the right side.

 

[Deleted member 4993]

Hi,
I'm working on calculus problems, and there is one equation I can't differentiate. It's this one:
View attachment 28384
Because I've done it using the power and the chain rule, I got the following:
View attachment 28385
But the app I use to verify the answers constantly gives this one and only answer:
View attachment 28386
I'm completely lost.
What is the derivative of the function? And do you know why the computer answers v' = 0 ?
Thank you,
Air Conditioner

I think your app is trying to calculate \(\displaystyle \frac{dV}{dx}\), where as the given V is NOT a function of 'x'. So V is "constant" with respect to 'x'- thus

V' = \(\displaystyle \frac{dV}{dx}\) = 0

 

[JeffM]

Your app flunks.

[math]V = 2 \pi r^ 2 \sqrt{R^2 - r^2} = uv, \text { where } u = 2 \pi r^2 \text { and } v = \sqrt{R^2 - r^2} \implies[/math]
[math]\dfrac{dV}{dx} = u * \dfrac{dv}{dx} + v * \dfrac{du}{dx} \implies \text {WHAT?}[/math]

 

[Air conditionner]

Okay, thank you, so it's not v' = 0.
But, pka,

Lets play the back-of-the-book game. Look up the answer to this and see:
[imath]D_r\left(2\pi r^2\sqrt{R^2-r^2}\right)=\dfrac{4\pi r R^2-6\pi r^3}{\sqrt{R^2-r^2}}[/imath] SEE THIS

Now your job is to find out why is that the answer?

I would very much like to understand this. Could you please reexplain?

Your app flunks.

[math]V = 2 \pi r^ 2 \sqrt{R^2 - r^2} = uv, \text { where } u = 2 \pi r^2 \text { and } v = \sqrt{R^2 - r^2} \implies[/math]
[math]\dfrac{dV}{dx} = u * \dfrac{dv}{dx} + v * \dfrac{du}{dx} \implies \text {WHAT?}[/math]

So my answer is probably correct, thank you.

 

[topsquark]

So my answer is probably correct, thank you.

Ummm... no.
[imath]\dfrac{d}{dr} \left ( 2 \pi r^2 \sqrt{R^2 - r^2} ~ \right )[/imath]

[imath]= \dfrac{d}{dr} \left ( 2 \pi r^2 \right ) \cdot \sqrt{R^2 - r^2} + 2 \pi r^2 \cdot \dfrac{d}{dr} \left ( \sqrt{R^2 - r^2} ~ \right )[/imath]

[imath]= 2 \pi (2r) \cdot \sqrt{R^2 - r^2} + 2 \pi r^2 \cdot \dfrac{1}{2} \dfrac{1}{\sqrt{R^2 - r^2}} \cdot (-2r)[/imath]

[imath]= 4 \pi r \sqrt{R^2 - r^2} - \dfrac{2 \pi r^3}{\sqrt{R^2 - r^2}}[/imath]

Then just add the fractions.

Is this enough or do you need help with the chain rule in the second term?

-Dan

 

[Air conditionner]

Ummm... no.
[imath]\dfrac{d}{dr} \left ( 2 \pi r^2 \sqrt{R^2 - r^2} ~ \right )[/imath]

[imath]= \dfrac{d}{dr} \left ( 2 \pi r^2 \right ) \cdot \sqrt{R^2 - r^2} + 2 \pi r^2 \cdot \dfrac{d}{dr} \left ( \sqrt{R^2 - r^2} ~ \right )[/imath]

[imath]= 2 \pi (2r) \cdot \sqrt{R^2 - r^2} + 2 \pi r^2 \cdot \dfrac{1}{2} \dfrac{1}{\sqrt{R^2 - r^2}} \cdot (-2r)[/imath]

[imath]= 4 \pi r \sqrt{R^2 - r^2} - \dfrac{2 \pi r^3}{\sqrt{R^2 - r^2}}[/imath]

Then just add the fractions.

Is this enough or do you need help with the chain rule in the second term?

-Dan

No, this is enough,
Thank you very much!