# DE book does not justify this claim

## Homework Statement

My DE book claims that

$\frac{d\mu(t)/dt}{\mu(t)}$

is equivalent to

$\frac{d}{dt}ln|\mu(t)|$

But says nothing to justify this claim.

Can you?

Related Calculus and Beyond Homework Help News on Phys.org
gabbagabbahey
Homework Helper
Gold Member
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 $\frac{d}{dx} \ln|x| = \frac{1}{x}$.

I see how from the bottom, the top follows by direct application of the chain rule.

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

Bacle2

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).

gabbagabbahey
Homework Helper
Gold Member
I see how from the bottom, the top follows by direct application of the chain rule.

But given the top, I don't know how to come to the bottom expression.
Just compute the anti-derivative:

$$\int \frac{ \frac{d\mu}{dt} }{ \mu(t) }dt = \int \frac{d\mu}{\mu} = \ln|\mu| + C$$

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:

Remember that:
$\large \frac{δ}{δt}ln|x|=\frac{1}{x}$

but since you have a another function inside the log, you need to apply the chain rule, so:
$\large \frac{δ}{δt}ln|\mu(t)|=\frac{1}{\mu(t)}\cdot\frac{δ\mu(t)}{δt}$

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:

$\frac{d\mu(t)/dt}{\mu(t)} = q$

Integrating both sides

$ln|\mu(t)| = qt$

Then taking the derivative of both sides

$\frac{dln|\mu(t)|}{dt} = q$

Therefore:

$\frac{d\mu(t)/dt}{\mu(t)} = \frac{dln|\mu(t)|}{dt} = q$

Is satisfactory.

vela
Staff Emeritus
Homework Helper
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:

$\frac{d\mu(t)/dt}{\mu(t)} = q$

Integrating both sides

$ln|\mu(t)| = qt$
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.

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

gabbagabbahey
Homework Helper
Gold Member
q is constant, I just put it there so I could visualize it thoroughly.
But you defined $q$ as being equal to $\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}$, so it is only constant if $\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}$ is constant, which it isn't, in general.

It is constant, here.

If it weren't, instead of qt it would be an arbitrary integral notation of q. It doesn't matter.

#### Attachments

• 12 KB Views: 351
gabbagabbahey
Homework Helper
Gold Member

though I don't know why they dropped the absolute value
Try computing $\frac{d}{dx}\ln(x)$ when $x<0$. There's a reason that calculus books give the antiderivative of $\frac{1}{x}$ as $\ln|x|+C$ instead of $\ln(x)+C$.