Question

Find the differential of the function y=x3sin(9x).

Answer 1

\[\Large y=x^3\sin9x\]And we're supposed to find the differential dy?

Answer 2

dx

Answer 3

dx? :o hmm ok..\[\Large dy=\color{royalblue}{(x^3)'}\sin9x+x^3\color{royalblue}{(\sin9x)'}\]We'll need to apply the product rule.
The blue portions we'll need to take derivatives of.
Understand the setup? :o

Answer 4

yes thank you

Answer 5

3x^2sin9x+x^3cos9x would be right?

Answer 6

Woops! When you take the derivative of the sin9x, don't forget your chain rule!! :O
We should get an extra factor of 9, yes?\[\Large (\sin9x)' \quad=\quad \cos9x(9x)'\quad=\quad 9\cos9x\]Rest of it looks good though.

Answer 7

wait i am confused

Answer 8

chain rule confusing?

Answer 9

yea

Answer 10

So like.. here's the simpler explanation.
We can go with something more formal if this doesn't make sense.
Ummmmmm...

Chain rule tells us to do a couple things when we take a derivative.
Make a copy of the inner function, and multiply by its derivative.
Example:\[\Large \frac{d}{dx}(3x-2)^2 \quad=\quad 2(3x-2)\color{royalblue}{\frac{d}{dx}(3x-2)}\]
\[\Large =2(3x-2)\color{royalblue}{(3)}\]

Answer 11

So in that example ^ we applied the power rule to the outermost function (stuff)^2.
Then we multiplied by the derivative of the "stuff"

Answer 12

If you want the more formal definition, you can think of it as a composition of functions.
Chain rule says this:\[\Large \left[f(g(x))\right]' \quad=\quad f'(g(x))\cdot g'(x)\]

So in our problem we can think of it like this:\[\Large f(x)=\sin x, \qquad g(x)=9x\]
\[\Large f(g(x))=\sin9x\]

Answer 13

\[\Large f'(g(x))=\cos9x,\qquad g'(x)=9\]

Answer 14

Chain rule can be a tough one to get a good grasp on, even harder to explain lol :(

Answer 15

thank you for helping me out

Answer 16

np c:

Answer 17

@zepdrix ' i liked your way of clearing the concepts. .