- #1

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).

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

So, differentiating both sides with respect to

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

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:

[tex]x^2 = x => x = 1[/tex]

Differentiating both sides:

[tex]2 x = 1 => x = \frac{1}{2}[/tex]

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?

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

So, differentiating both sides with respect to

*r*and using the Fundamental Theorem:[tex]\rho(r) * 4 \pi r^2 = \frac{6 Q r^5}{R^6}[/tex]

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:

[tex]x^2 = x => x = 1[/tex]

Differentiating both sides:

[tex]2 x = 1 => x = \frac{1}{2}[/tex]

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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