# Isaac Physics 'Implicit Differentiation 2'

Watch
Announcements

https://isaacphysics.org/questions/i...9-7215847615b0

I understood Implicit Differentiation for when you just had an equation in terms of, for instance, x and y, but this stuff I don't get.

Could someone explain to me how'd you approach RT when evaluating it for d/dV. Would it be T*dR/dV ?? I am confused.

I understood Implicit Differentiation for when you just had an equation in terms of, for instance, x and y, but this stuff I don't get.

Could someone explain to me how'd you approach RT when evaluating it for d/dV. Would it be T*dR/dV ?? I am confused.

0

reply

Report

#2

I don’t even do physics but I think I know this

Just simply expand and differentiate

T is a constant

Just simply expand and differentiate

T is a constant

Last edited by CaptainDuckie; 2 weeks ago

0

reply

(Original post by

I don’t even do physics but I think I know this

Just simply expand and differentiate

T is a constant so it’ll be just R on the right

**CaptainDuckie**)I don’t even do physics but I think I know this

Just simply expand and differentiate

T is a constant so it’ll be just R on the right

**implicit**differentiation question, read up on the concept and you'll see that it isn't just the normal kind of differentiation you'd do in Year 12.

0

reply

Report

#4

(Original post by

This is an

**domm1**)This is an

**implicit**differentiation question, read up on the concept and you'll see that it isn't just the normal kind of differentiation you'd do in Year 12.Implicit differentiate it, expand brackets, then use R as a constant on the right.

0

reply

(Original post by

Yes, I’m not in year 12.

Implicit differentiate it, expand brackets, then use R as a constant on the right.

**CaptainDuckie**)Yes, I’m not in year 12.

Implicit differentiate it, expand brackets, then use R as a constant on the right.

0

reply

Report

#6

(Original post by

But I thought that I'd be evaluating d/dV for each term, so for RT it would be 'd/dV*RT' which is that same as saying dRT/dV, but we're evaluating w.r.t V not R so what do I do? Apologies that my attempt to explain my confusion is all over the place.

**domm1**)But I thought that I'd be evaluating d/dV for each term, so for RT it would be 'd/dV*RT' which is that same as saying dRT/dV, but we're evaluating w.r.t V not R so what do I do? Apologies that my attempt to explain my confusion is all over the place.

0

reply

(Original post by

Write out what you think it is

**CaptainDuckie**)Write out what you think it is

0

reply

(Original post by

I am quite clueless on this question so I'm really not sure, hence why I put this post out.

**domm1**)I am quite clueless on this question so I'm really not sure, hence why I put this post out.

0

reply

(Original post by

It's not the entire expression which confuses me when differentiating, but certain terms, such as RT and -pb. How would you differentiate those terms w.r.t V? This is where my confusion lies.

**domm1**)It's not the entire expression which confuses me when differentiating, but certain terms, such as RT and -pb. How would you differentiate those terms w.r.t V? This is where my confusion lies.

0

reply

Report

#12

**domm1**)

It's not the entire expression which confuses me when differentiating, but certain terms, such as RT and -pb. How would you differentiate those terms w.r.t V? This is where my confusion lies.

0

reply

Report

#13

(Original post by

When differentiating -pb w.r.t V you would use the product rule, such that d(-pb)/dV = -p*db/dV + b*d(-p)/dV , right?

**domm1**)When differentiating -pb w.r.t V you would use the product rule, such that d(-pb)/dV = -p*db/dV + b*d(-p)/dV , right?

I think R would just be R

0

reply

Report

#14

**domm1**)

When differentiating -pb w.r.t V you would use the product rule, such that d(-pb)/dV = -p*db/dV + b*d(-p)/dV , right?

0

reply

(Original post by

Yes, although it would be *way* less confusing to look at - d(pb)/dV

**DFranklin**)Yes, although it would be *way* less confusing to look at - d(pb)/dV

d/dV*RT = T (from advice from CaptainDuckie)

d/dV*pV = p

d/dV*aV^-1 = -aV^-2

-d/dV*pb = -(p*db/dV + b*dp/dV)

-d/dV *abV^-2 = 2abV^-3 ?

0

reply

Report

#16

(Original post by

...

**domm1**)...

**Edit:**It also gives a "correct" answer when you treat it as such.

0

reply

(Original post by

From what I recall of physics, R is just a constant, so if T is as well, then the entire right hand side is simply a constant.

**ghostwalker**)From what I recall of physics, R is just a constant, so if T is as well, then the entire right hand side is simply a constant.

**Edit:**It also gives a "correct" answer when you treat it as such.
0

reply

Report

#18

(Original post by

I didn't think R was a constant?

**domm1**)I didn't think R was a constant?

0

reply

Report

#19

(Original post by

So if I were to evaluate each term I should get: (?)

d/dV*RT = T (from advice from CaptainDuckie)

d/dV*pV = p

d/dV*aV^-1 = -aV^-2

-d/dV*pb = -(p*db/dV + b*dp/dV)

-d/dV *abV^-2 = 2abV^-3 ?

**domm1**)So if I were to evaluate each term I should get: (?)

d/dV*RT = T (from advice from CaptainDuckie)

d/dV*pV = p

d/dV*aV^-1 = -aV^-2

-d/dV*pb = -(p*db/dV + b*dp/dV)

-d/dV *abV^-2 = 2abV^-3 ?

Because its just like saying two constants multiplied by each other would differentiate to 0

0

reply

X

### Quick Reply

Back

to top

to top