<p>can someone help me with a derivation for the chain rule for differentiation??? thanks much</p>
<p>Monsieur Google would be helpful.</p>
<p>post some examples, problems...</p>
<p>it think its</p>
<p>F(H(x))</p>
<p>chain = F'(H(x))H'(X)</p>
<p>you guys don't understand his question. he wants to know WHY the chain rule is f'(h(x)) h'(x) How was this rule derived/discovered from other things? Ya know?? That can probably be found with persistence using Googlie Wooglie</p>
<p>differentiation... so it will be something like
1/. cosx + tany^2 = x
2/. sin2xy + yx^2 = 2y</p>
<p>Yes, that would be what differentiation is, but he wants HOW / FROM WHAT the chain rule was described hence his question (at least I think)</p>
<p>so... okinawa, it's UR turn to solve it, lol. I just made up those problems... so try it</p>
<p>well i can already tell you we didn't learn derivatives with an x and a y in them. just with xs so I can't solve those. we just finished chain rule and we just wasted like the past 2 weeks doing packets that he collected for test grades on stuff we learned. so i don't know those and this thread is one in which the OP wanted to know how he could derive the chain rule..like a proof</p>
<ol>
<li>dy/dx=(sinx+1)/(2y(sec(y^2)^2))</li>
<li>dy/dx=(2ycos(2xy)+2xy)/(2-x^2-2xcos(2xy))</li>
</ol>
<p>ok fine fine!!! budy! lol ^_^
1/. cosx + tany^2 = x
derivative: -sinx + (2y)(dy/dx)(secy^2)^2 = 1
solve for dy/dx, we get:
dy/dx = (1+sinx)/(2y(secy^2)^2)
2/. sin2xy + yx^2 = 2y
derivative: (cos2xy)(2xdy/dx + 2y) + (x^2)dy/dx + 2xy = 2dy/dx
(2x)(dy/dx)(cos2xy) + 2ycos2xy + (x^2)(dy/dx) + 2xy = 2(dy/dx)
put all dy/dx on one side and then take dy/dx out
dy/dx (2xcos2xy + x^2 - 2) = -2xy - 2ycos2xy
solve for dy/dx
dy/dx = (-2xy - 2ycos2xy)/(2xcos2xy + x^2 -2)
I just made up these 2 problems, so it seems to be really confusing! but the idea is just the same as chain rule for normal function as y=sinx^2 . But for the chain rule of differentiation, everytime you take the derivative of y, remember to put dy/dx.
Just as for y=sinx^2
when we take the derivative of this function, we are actually taking the derivative of both side. So derivative of the left side is dy/dx, and derivative of the right side is 2xcosx^2. Therefore, dy/dx = 2xcosx^2.
If you still dont get it, here's another example, I'll make it simple
y^2 = sinx
derivative of the left side: 2y (dy/dx)
derivative of the right side: cosx (dx/dx), but dx/dx = 1, therefore we never write them when taking derivatives.
so we have 2y (dy/dx) = cosx
--> dy/dx = (cosx) / (2y)</p>
<p>if you still dont get it, lemme know...</p>
<p>NO NO NO</p>
<p>this is what I think he wants</p>
<p>here's a better one at the bottom</p>
<p>arggg!!!! you should've told me that it's a pdf file... I opened it, and caused my computer froze >_<, it's allllll your fault, jk</p>
<p>hardy har har..well did you see it? do you have dialup?</p>
<p>I have DSL, but for some reasons my computer cant open any pdf file. grrr!!!! hate it!</p>
<p>Well, some symbols may be odd or off etc because of copying off pdf, but here is sort of what it means:</p>
<p>Derivation of the Chain Rule
Suppose y = f g(x). Assuming f and g have derivatives where appropriate, the Chain Rule says that
(f g)0 = (f0 g) </p>
<p>yeah okinawa u were right...i needed a "proof" of the chain rule..like how to derive it from the difference quotient or something..thanks okinawa..and everyone else who came up with problems?</p>