# Need a good definition of derivatives **URL:** \ **Category:** Factual Questions\ **Created:** [May 12, 2002, 11:08pm UTC](https://boards.straightdope.com/t/need-a-good-definition-of-derivatives/108702 "2002-05-12T23:08:20Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 17, 2002, 7:11pm UTC](https://boards.straightdope.com/t/need-a-good-definition-of-derivatives/108702/21 "2002-05-17T19:11:47Z")
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. ---
**Author:** ![Qwertyasdfg](https://avatars.discourse-cdn.com/v4/letter/q/ecc23a/32.png) [@Qwertyasdfg](https://boards.straightdope.com/u/Qwertyasdfg)\ **Post date:** [July 5, 2002, 2:09am UTC](https://boards.straightdope.com/t/need-a-good-definition-of-derivatives/108702/22 "2002-07-05T02:09:58Z")
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? ---
**Author:** ![tekgraf](https://avatars.discourse-cdn.com/v4/letter/t/8491ac/32.png) [@tekgraf](https://boards.straightdope.com/u/tekgraf)\ **Post date:** [July 5, 2002, 2:31am UTC](https://boards.straightdope.com/t/need-a-good-definition-of-derivatives/108702/23 "2002-07-05T02:31:50Z")
Jesus, give it a rest! ---
**Author:** ![g8rguy](https://avatars.discourse-cdn.com/v4/letter/g/7cd45c/32.png) [@g8rguy](https://boards.straightdope.com/u/g8rguy)\ **Post date:** [July 5, 2002, 3:27am UTC](https://boards.straightdope.com/t/need-a-good-definition-of-derivatives/108702/24 "2002-07-05T03:27:07Z")
> [@](#): > > 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 = 2x_cos(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(A_sin(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 A_cos(z)w = Aw_cos(wt) ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [July 5, 2002, 3:38am UTC](https://boards.straightdope.com/t/need-a-good-definition-of-derivatives/108702/25 "2002-07-05T03:38:20Z")
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 [Previous page](https://boards.straightdope.com/t/need-a-good-definition-of-derivatives/108702.md?page=1)