- #1

1MileCrash

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## Homework Statement

My DE book claims that

[itex]\frac{d\mu(t)/dt}{\mu(t)}[/itex]

is equivalent to

[itex]\frac{d}{dt}ln|\mu(t)|[/itex]

But says nothing to justify this claim.

Can you?

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- Thread starter 1MileCrash
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- #1

1MileCrash

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My DE book claims that

[itex]\frac{d\mu(t)/dt}{\mu(t)}[/itex]

is equivalent to

[itex]\frac{d}{dt}ln|\mu(t)|[/itex]

But says nothing to justify this claim.

Can you?

- #2

gabbagabbahey

Homework Helper

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But says nothing to justify this claim.

Why would it? Most Authors would assume that if a student is studying differential equations, they have already mastered the basics of single variable calculus, such as computing derivatives and anti-derivatives.

This is just an application of the chain rule, combined with the fact that [itex]\frac{d}{dx} \ln|x| = \frac{1}{x}[/itex].

- #3

1MileCrash

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But given the top, I don't know how to come to the bottom expression.

- #4

Bacle2

Science Advisor

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But that seems just like the chain rule , (though I don't know why they dropped the

absolute value): d/dt f(g(t))= f'(g(t))g'(t) , where f(t)=ln|t| g(t)=u(t). Or

maybe you can do a triple chain rule f(g(h)) , with g(t)=|t|, h(t)=u(t), and

f(t)=ln(t).

- #5

gabbagabbahey

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But given the top, I don't know how to come to the bottom expression.

Just compute the anti-derivative:

[tex]\int \frac{ \frac{d\mu}{dt} }{ \mu(t) }dt = \int \frac{d\mu}{\mu} = \ln|\mu| + C[/tex]

More importantly though, given how common a function the natural logarithm is, you need to be able to recognize its derivative when you see it. A big part of problem solving with differential equations requires you to have experience with calculating common derivatives (which is why most calculus textbooks have all those practice problems in them ) and applying that experience, by recognizing derivatives of common functions when you see them.

Last edited:

- #6

Nessdude14

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Remember that:

[itex]\large \frac{δ}{δt}ln|x|=\frac{1}{x}[/itex]

but since you have a another function inside the log, you need to apply the chain rule, so:

[itex]\large \frac{δ}{δt}ln|\mu(t)|=\frac{1}{\mu(t)}\cdot\frac{δ\mu(t)}{δt}[/itex]

- #7

1MileCrash

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Nonetheless:

[itex]\frac{d\mu(t)/dt}{\mu(t)} = q[/itex]

Integrating

[itex]ln|\mu(t)| = qt[/itex]

Then taking the derivative of

[itex]\frac{dln|\mu(t)|}{dt} = q[/itex]

Therefore:

[itex]\frac{d\mu(t)/dt}{\mu(t)} = \frac{dln|\mu(t)|}{dt} = q[/itex]

Is satisfactory.

- #8

vela

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This all seems kind of pointless. Integrating ##\frac{d\mu(t)/dt}{\mu(t)}## to get ##\ln|\mu(t)|##, you've used the fact you're trying to show, so why bother? Also, the righthand side is only equal to qt if q is a constant, which it most likely isn't.That's not really my issue. I just like to see it written out even if it is recognizable - I can integrate and differentiate until the cows come home.

Nonetheless:

[itex]\frac{d\mu(t)/dt}{\mu(t)} = q[/itex]

Integratingboth sides

[itex]ln|\mu(t)| = qt[/itex]

- #9

1MileCrash

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q is constant, I just put it there so I could visualize it thoroughly.

- #10

gabbagabbahey

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q is constant, I just put it there so I could visualize it thoroughly.

But you defined [itex]q[/itex] as being equal to [itex]\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}[/itex], so it is only constant if [itex]\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}[/itex] is constant, which it isn't, in general.

- #11

- #12

gabbagabbahey

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though I don't know why they dropped the absolute value

Try computing [itex]\frac{d}{dx}\ln(x)[/itex] when [itex]x<0[/itex]. There's a reason that calculus books give the antiderivative of [itex]\frac{1}{x}[/itex] as [itex]\ln|x|+C[/itex] instead of [itex]\ln(x)+C[/itex].

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