Question

help

Answer 1

take the derivative ... what have you attempted?

Answer 2

we have a function inside of another function .... any rules come to mind?

Answer 3

if we let u = 5x^2

how would you process cos(u) ?

Answer 4

\[y=f(g(x))\]
\[\frac{dy}{dx}=\frac{d[f(g(x)]}{d[g(x)]}\]
the tops equivalent, but we dont have dx on the bottom. if we were to multiply the right side by some fraction looking thing that would cancel out the d[g(x)] that would be great ...
\[\frac{dy}{dx}=\frac{d[f(g(x)]}{d[g(x)]}~\frac{d[g(x)]}{dx}\]
now its better since d[g(x)] cancel out: to make it easier to see, less clutter. let g(x) = u
\[\frac{dy}{dx}=\frac{d[f(g(x)]}{du}~\frac{du}{dx}\]
\[\frac{dy}{dx}=\frac{dy}{du}~\frac{du}{dx}\]

Answer 5

this is called the chain rule

Answer 6

in this case:

y = cos(u) ; and u = 5x^2

what is dy/du?
what is du/dx?

Answer 7

what about it is confusing? and what are you refering to by 'd' ?

Answer 8

when we differentiate a function with respect to some variable (*) the differentation operator is notated as: d/(d*)

Answer 9

ohh

Answer 10

when we derive a function y wrt.x we find d(y)/dx

when we derive a function y wrt.u we find d(y)/du

when we derive a function f wrt.t we find d(f)/dt

does this make sense?

Answer 11

if we let y be a function of u; then dy/du should be easily determined

if y = cos(u) what would you say dy/du (the derivative) is?

Answer 12

the derivative of cos(u) is -sin(u) ... dy/du = -sin(u)
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if we let u be a function of x, then du/dx should be easily deternined

if u = 5x^2, then du/dx (the derivative) is?

Answer 13

my personal belief is that they teach derivatives backwards.

you get stuck in a mindset that there is some special magical properties of the name 'x' that makes things work. its just not so. the chain rule is always working underneath the operation.

Answer 14

when starting our with y as a function of x, they fail to tell you that:
\[\frac{dy}{dx}=\frac{dy}{dx}~\frac{dx}{dx}\]
the dont mention this simply because dx/dx = , which is the most fundamental property of fractions (excluding of course 0/0)

Answer 15

dx/dx = 1

Answer 16

when we have a function that is made up of other functions, the chain rule becomes less trivial and more pronounced

\[\frac{dy}{dx}=\frac{dy}{da}\frac{da}{db}\frac{db}{d*}...\frac{d*}{dx}\]

Answer 17

yeah, thats a site issue that is just all to common

Answer 18

dy/5x^2

Answer 19

\[y=cos(5x^2)\]
now, with respect fo 5x^2
\[\frac{dy}{d[5x^2]}=-sin(5x^2)\]
but we dont want this with respect to 5x^2, we only want it with respect to x
so multiply both sides by d(5x^2)/dx
\[\frac{dy}{\cancel{d[5x^2]}}~\frac{\cancel{d[5x^2]}}{dx}=-sin(5x^2)*10x=\frac{dy}{dx}\]

Answer 20

where was 10x taken from?

Answer 21

ohh! got it!

Answer 22

what is the derivative of 5x^2 with respect to x? yeah

Answer 23

it really is simple math :) good luck