Help! Calc! problem attached

Question

voltron21 · Answer

attachment

psymon · Answer

Is that the main confusion, why pi is there?

voltron21 · Answer

no pi is irrelevant to the problem...its a tool box

voltron21 · Answer

which ever is easier

agent0smith · Answer

If you write it like this, you can use the chain rule$$\Large P = V^2 R (R + r)^{-2}$$
Remember... V^2R is just a constant, so just leave it where it is.

psymon · Answer

Oh, duh, yeah, just do chain rule, im stupid xD

voltron21 · Answer

Ur not stupid...so guide me through?

agent0smith · Answer

Let's do it for a simpler function... $$\Large y = (1+x)^{-2}$$ can you find the derivative of this?

psymon · Answer

Yeah,Ill let smith take over, I feel like Ill just complicate things, sorry xD

voltron21 · Answer

my brain just shut off

voltron21 · Answer

i know to use the power rule

voltron21 · Answer

do i bring down the exponent then multiply through?

voltron21 · Answer

@agent0smith ???

agent0smith · Answer

Bring down the exponent, decrease the exponent by 1, and multiply by the derivative of what's inside the brackets.

voltron21 · Answer

so -2(0)^-1
0....because (x+1)'= 0...right?

voltron21 · Answer

oh no wait thats super wrong

agent0smith · Answer

You didn't decrease the power by 1, and the function inside the brackets should stay as is, only the power on it decreases. And the derivative of (x+1) is not zero.

Yep, super wrong :P

voltron21 · Answer

what does this have to do with the problem i asked about?

agent0smith · Answer

It's almost exactly the same, only with constants, which we can put back in later.

agent0smith · Answer

All we'll have to do is replace them at the end.

voltron21 · Answer

ok so i can do $$1/(x+2)^2$$

voltron21 · Answer

i mean (x+1)

voltron21 · Answer

y'= $$\frac{ -2 }{ (x+1)^3 }$$

voltron21 · Answer

How does that apply to the other problem?

agent0smith · Answer

Because as i said, all you have to do is put the constants back in.

voltron21 · Answer

could u show me?

agent0smith · Answer

Just put them back in, where they were before. From here you should be able to tell where they go.

$$\Large P = \frac{ V^2 R  }{(R + r)^{2}}$$
$$\Large y = \frac{ 1 }{ (1+x) ^2 }$$

voltron21 · Answer

so do i just replace r with x?

agent0smith · Answer

Yes. And don't forget the other constants.

agent0smith · Answer

$$\Large y = \frac{ -2 }{ (x+1)^3 }$$ put them all in here.

voltron21 · Answer

is it 1/(1+r)^2 then i multiply by P and find the derivative of that?

agent0smith · Answer

No... you already have the derivative. Just put the constants back where they belong. Put the V^2R back on the numerator, and the R back in the denominator.

agent0smith · Answer

And change y into P.

voltron21 · Answer

P=-2V^2R/(R+1)^3

agent0smith · Answer

You're missing the r in the denominator.

We changed 
R+r
into 
1+x

voltron21 · Answer

oh ok thanks so much for being patient with me :)

agent0smith · Answer

The reason i showed you that way is because the constants don't really affect anything. Just ignore them, leave them where they are.