Need a good definition of derivatives

[Chronos]

The chain rule is much simpler, by the way, in Leibnitz notation. There, it’s just dz/dx = (dz/dy) * (dy/dx) , exactly like you’d guess if those were fractions.

[Qwertyasdfg]

Ok, I have more questions, which I’ll put here rather than start a new thread:

1. When taking the derivative of a trig function, if the degrees part is a fuction (for example sin (x^2) ) should one differentiate the internal function too? In other words, would that be cos (x^2) or cos (2x) ?

2. Here’s a practice question from the book I’m using. I know the answer (from the book), but I dont understand it. (This isn’t a home work problem; I’m teaching myself calculus to be prepared for some courses I’m taking next school year.)
   V = A sin (wt). A and w are constants. Find dV/dt.
   Here were my thoughts (I got the wrong answer.)

First, make the following statements:
u = sin (wt)
z = wt
According to chain rule (I have no idea what the notation should be, I’ve called derviatives d/dx for now, please correct me on this):
dV/dt = (d/dx Au)(d/dx sin z)(d/dx wt)
The derviative of a constant times a variable is the constant times the dervative of the variable, so:
d/dx Au = A d/dx u
d/dx wt = w d/dx t
The derivative of a sine is a cosine so:
d/dx sin z = cos z.
So you get: (A)(d/dx u)(cos z)(w)(d/dx t)
Plugging in for u and z:
Aw(d/dx sin wt)(cos wt)(d/dx t)
Which comes out to: Aw cos[sup]2/sup(d/dx t)
The right answer however, is: wa cos(wt)
What did I do wrong?

[tekgraf]

Jesus, give it a rest!

[g8rguy]

1. When taking the derivative of a trig function, if the degrees part is a fuction (for example sin (x^2) ) should one differentiate the internal function too? In other words, would that be cos (x^2) or cos (2x) ?

Using the chain rule with u=x[sup]2[/sup], we’d have [sup]d sin(u)[/sup]/[sub]dx[/sub] = [sup]d sin(u)[/sup]/[sub]du[/sub][sup]du[/sup]/[sub]dx[/sub] = cos(u)[sup]du[/sup]/[sub]dx[/sub] = cos(u)2x = 2xcos(x[sup]2[/sup]). So yes, differentiate the inside part, but that gets pulled out of the trig function. Generally, whenever you see a composite function like this, you’ll have to use the chain rule.

V = A sin (wt). A and w are constants. Find dV/dt.
Here were my thoughts (I got the wrong answer.)

First, make the following statements:
u = sin (wt)
z = wt
According to chain rule (I have no idea what the notation should be, I’ve called derviatives d/dx for now, please correct me on this):
dV/dt = (d/dx Au)(d/dx sin z)(d/dx wt)
The derviative of a constant times a variable is the constant times the dervative of the variable, so:
d/dx Au = A d/dx u
d/dx wt = w d/dx t
The derivative of a sine is a cosine so:
d/dx sin z = cos z.
So you get: (A)(d/dx u)(cos z)(w)(d/dx t)
Plugging in for u and z:
Aw(d/dx sin wt)(cos wt)(d/dx t)
Which comes out to: Aw cos[sup]2/sup(d/dx t)
The right answer however, is: wa cos(wt)
What did I do wrong?

Basically, you made the entire thing far too complicated and got confused about the chain rule. First of all, note that differentiation with respect to variable p is d/dp; that’s part of your confusion. So let’s call wt=z, like you did. You want, then,
[sup]d(Asin(z))[/sup]/[sub]dt[/sub] = A[sup]d(sin(z))[/sup]/[sub]dz[/sub]*[sup]dz[/sup]/[sub]dt[/sub]

The derivative of sin(z) is cos(z), the derivative of wt is w, so plugging these back in we get Acos(z)w = Awcos(wt)

[Ring]

V = A sin (wt).

Instead of letting u = sin (wt) let u = wt

So du / dt = w

Now dV / dt = A cos (u) du / dt = A cos (wt) * w