# Differentiating the equation for the mechanical energy of a spring

Callumnc1
Homework Statement:
Relevant Equations:
## E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2 ##
Why when we differentiate ## E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2 ## with respect to time the answer is ## \frac {dE}{dt} = mva + kxv ##?

I though it would be ##\frac {dE}{dt} = ma + kv ##.

Many thanks!

Homework Helper
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Relevant Equations:: ## E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2 ##

Why when we differentiate ## E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2 ## with respect to time the answer is ## \frac {dE}{dt} = mva + kxv ##?

I though it would be ##\frac {dE}{dt} = ma + kv ##.

Many thanks!
What is the derivative of ##x^2## wrt ##x##?

Callumnc1
Gold Member
Chain rule. x and v are functions of t.

Callumnc1
Gold Member
The time rate of change in energy is power.

With what you thought it would be the units on the RHS are Force added to Force per unit time. Not only are neither of them power, they are also dimensionally inconsistent with each other.

Last edited:
Callumnc1
Callumnc1
What is the derivative of ##x^2## wrt ##x##?
Thank you for your reply @haruspex ! I don't think I've ever taken the derivative of ##x^2## wrt ##x##. I think I've only the derivative of ##y## wrt ##x##. How would I take the derivative?

Callumnc1
Chain rule. x and v are functions of t.

Frabjous
Callumnc1
The time rate of change in energy is power.

With what you thought it would be the units on the RHS are Force added to Force per unit time. Not only are neither of them power, they are also dimensionally inconsistent with each other.

erobz
Gold Member
Thank you for your reply @haruspex ! I don't think I've ever taken the derivative of ##x^2## wrt ##x##. I think I've only the derivative of ##y## wrt ##x##. How would I take the derivative?
You almost certainly have, you just don’t realize it. ##y=x^2##

Callumnc1
Callumnc1
You almost certainly have, you just don’t realize it. ##y=x^2##

Oh I thought that was taking the derivative of y with respect to x to get ##2x##?

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Oh I thought that was taking the derivative of y with respect to x to get ##2x##?
That’s correct. Then you apply the chain rule. First differentiate ##y =x^2 ## wrt ##x##, then ##x## wrt ##t##.

Callumnc1
Callumnc1
That’s correct. Then you apply the chain rule. First differentiate ##y =x^2 ## wrt ##x##, then ##x## wrt ##t##.
Thank you for your reply @erobz! I think it would be ## y = (2x)\frac {dx}{dt} ##

erobz
Gold Member
Thank you for your reply @erobz! I think it would be ## y = (2x)\frac {dx}{dt} ##
Do you see how it works out?

Callumnc1
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Thank you for your reply @erobz! I think it would be ## y = (2x)\frac {dx}{dt} ##
Not quite. You must do the same to each side of an equation.
The derivative of y wrt x is ##\frac{dy}{dx}##.
The derivative of ##x^2## wrt x is ##\frac{d(x^2)}{dx}=2x##.
So differentiating both sides of ##y=x^2## wrt x gives
##\frac{dy}{dx}=2x##.

Last edited:
Callumnc1
Gold Member
Not quite. You must do the same to each side of an equation.
The derivative of y wrt x is ##\frac{dy}{dx}##.
The derivative of ##x^2## wrt x is ##\frac{d(x^2)}{dx}=2x##.
So differentiating both sides of ##y=x^2## wrt x gives
##\frac{dy}{dx}=x^2##.
last line typo:

$$\frac{dy}{dx}= 2x$$

Callumnc1 and haruspex
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last line typo:

$$\frac{dy}{dx}= 2x$$
thanks - corrected,

Callumnc1
Callumnc1
Not quite. You must do the same to each side of an equation.
The derivative of y wrt x is ##\frac{dy}{dx}##.
The derivative of ##x^2## wrt x is ##\frac{d(x^2)}{dx}=2x##.
So differentiating both sides of ##y=x^2## wrt x gives
##\frac{dy}{dx}=2x##.
Thank you for your replies @erobz and haruspex! Sorry, that was a silly mistake I should not have made!