1. Aug 18, 2011

### Alex1136

Hello everybody!

I am currently reading Feinman course of physics and there is one question about derivative -

I attached the formula. How did they got this derivative? I cannot understand this transformation, please help) Maybe smb could give a link where it is shown a rule how to get derivative in such case- A pity that I am not smart enough to solve this problem myself, it makes no fun when there are places in the book where I am not 100% sure what did the author meant.

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2. Aug 18, 2011

### Pengwuino

It's just the chain rule. Remember that $v = v(t)$ so it's just like taking the derivative ${{d}\over{dt}}\left(f^2(t)) \right) = 2f(t) f'(t)$

3. Aug 18, 2011

### Andrew Mason

This is just an application of the chain rule. See: http://en.wikipedia.org/wiki/Chain_rule

AM

4. Aug 18, 2011

### Alex1136

Hello,

sorry ( I am not really smart) but I would write just = 2f(t)

my problem is I cannot get why I have to multiply also with f '(t)

I cannot find exact explanation that passes to this case in wiki(

5. Aug 18, 2011

### Pengwuino

It should be in your calculus text, it's the chain rule.

Pretend for a second you already knew the form of the velocity to be something trivial (and totally non-physical so don't assume you'll ever ever see the velocity written like this) as $v(t) = t^2$. So your kinetic energy is

$T = mv(t)^2/2$

So let's plug in what we know about v(t) and we find $T = mt^4/2$ is your kinetic energy. Do the time derivative and you get that it is $2mt^3$. So that's the answer we KNOW is correct.

Now, using the v(t) we have, you would only get $T = mt^2$ with your way of thinking, which is not what we know is true. You have to multiply by $v'(t) = 2t$ to get the correct answer.

6. Aug 19, 2011

### Alex1136

Thank you very much for support Pengwuino! You explanation on example is perfect- now I got it.

I also found the rule:
d(ab)dx = a db/dx + b da/dx

I imagined it like d(v v)dx = v dv/dx + v dv/dx = 2v dv/dx

thank you and good weekend)))