Hi,

Could someone explain to me why the following step is allowed?
\(
e^{x}\frac{dy}{dx} + e^{x}y = e^{x}x
\)
\(
\frac{d}{dx}[ e^{x}y] = e^{x}x
\)

cheers,

 

The chain rule.

 

I think the chain rule has nothing to do here. It's just the expansion of the term
\((e^x \cdot y)'=(e^x)'\cdot y+e^x \cdot y'=e^x \cdot y+e^x \cdot y'\)

 

@georacer:

Risking sounding silly here: What do you mean "expaning of the term"?
I mean I am familiar with finding zeros by expanding term as in:
\(
x^{2}-9 = (x+3)\cdot(x-3)
\)
but I am not sure how this applies here.

Please elaborate.

 

i think the reason it is allowed is because its the same as multiplying fractions, for example

2/3*6 = (2*6)/3 = 12/3 = 4

think of the "d" part of the equation as just a function, just as if the equation was

-2/3*6 = -(2*6)/3 = -12/3 = -4

hope this was of some use

 

Nonononono! Nothing of the sort!

There is an identity in differential analysis that says that given two functions f and g which are differentiable, the followin is valid:
\(\frac{d}{dx}(f(x) \cdot g(x))=\frac{df(x)}{dx} \cdot g(x) + f(x) \cdot \frac{dg(x)}{dx}\)

Is that clear?

 

well ya helped me anyway, thank you!

 

Yes, thank you.

I have found where it came from (had to revert to manual for dummies) lol

If you take the derivative of: \(e^{x}y\)
i.e. \(\frac{d}{dx}[e^{x}y]\)

you get:
\(e^{x}\frac{dy}{dx}+e^{x}y\)

I was looking way to far out.

 

I think the chain rule has nothing to do here.

Georacer, I was going to respond angrily -- then I realized that my feeble brain mixed up the product rule with the chain rule. Thanks for correcting me. My only excuse is that I took calculus before most of you were born, so I can claim feeble-mindedness and forgetfulness. Of course, my wife would just claim it's overall incompetence.

 

Well, if your wife competes with you in math, then you have much more serious issues that mathematical mixups...

No harm done, be sure of that. The time will come when you will correct me in electronics, I 'm sure of that.

Have a nice evening!