Differentiating both sides of an equation

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SUMMARY

This discussion centers on the validity of differentiating both sides of an equation, specifically in the context of physics problems involving integrals. The example provided is q(r) = ∫₀ʳ ρ(s) * 4πs² ds = (Q r⁶)/(R⁶), where differentiation with respect to r yields ρ(r) * 4πr² = (6Q r⁵)/(R⁶). The key conclusion is that differentiation is valid when both sides of the equation represent a functional relationship that holds for all values of the variable, as opposed to equations like x² = x, which only hold true for specific values.

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  • Understanding of calculus, specifically differentiation and the Fundamental Theorem of Calculus.
  • Familiarity with integral equations and their applications in physics.
  • Knowledge of functional relationships and variable dependencies in mathematical expressions.
  • Basic algebraic manipulation skills to solve equations.
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awvvu
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When is it valid to differentiate both sides of an equation? I was working on a physics problem and came across this, where I had to solve for p(r).

q(r) = \int_0^r \rho(s) * 4 \pi s^2 ds = \frac{Q r^6}{R^6}

So, differentiating both sides with respect to r and using the Fundamental Theorem:

\rho(r) * 4 \pi r^2 = \frac{6 Q r^5}{R^6}

Solving for p(r), I get the right answer, so obviously this is what they expect me to do. What I'm wondering is why exactly is this valid, when this is not:

x^2 = x => x = 1

Differentiating both sides:

2 x = 1 => x = \frac{1}{2}

I think I came up with the gist of an explanation while typing this post up... but I'd really like a clear and more rigorous way to explain it. x^2 = x is only true for x = 1, so you can't differentiate both sides. But for my first equation, I assume that there is a p(r) that makes both sides of the equation true for all r. So, I can solve for this p(r). Am I right?
 
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The problem with your second equation is that it isn't one, at least in the same sense as your first equation. It uniquely determines what x is. Hence, x isn't really a variable - you know what it is - it's 1. It doesn't make sense to differentiate that expression then because x really isn't a variable in it.

In your first expression, you really do have a functional relationship between the integral and the other expression - it holds for any value of r. As a result, r is a variable and the expression can be differentiated on both sides of the equality.

So basically, the explanation you came up with was correct.
 
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Okay, thanks.
 

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