Question

Find the derivative
y=x^2 * cosx
y'=
Use y'=udv + vdu

Answer 1

Here is the setup for product rule.
\[\large y'=\color{royalblue}{(x^2)'}\cos x+x^2\color{royalblue}{(\cos x)'}\]

We need to take the derivative of the blue terms. That is what the primes indicate.
Understand the setup based on the rule they told you to use?

Answer 2

Yes I do!

Answer 3

Understand how to finish it? :) Gotta remember your trig derivatives!! :D

Answer 4

Don't you have to simplify it?

Answer 5

You have to `take the derivative` now.
All we did was setup the product rule so far.

Answer 6

Take the derivative of the blue terms. Those are the ones that need work still.

Answer 7

I'm afraid I'm a bit confused on finding the derivative :/

Answer 8

So we've successfully setup the product rule.

\[\large y'=\color{royalblue}{(x^2)'}\cos x+x^2\color{royalblue}{(\cos x)'}\]

Let's take it piece by piece.
We first want the derivative of x^2.

We can simply apply the power rule to find it.
Member the power rule? :O

Answer 9

Oooo is your picture a watermelon? XD Oh man that's fancy!

Answer 10

Refresh my memory please >_< sounds awfully familiar!

Answer 11

And yes it is!! :D I struggle just to cut one normally at times XD

Answer 12

Power Rule for Derivatives:
~Bring the power down as a coefficient in front.
~Decrease the power by 1.


Example:\[\large (x^3)' \qquad = \qquad 3x^{3-1} \qquad = \qquad 3x^2\]

Answer 13

OH YES. I remember it now!

Answer 14

So what would the derivative of \(\large x^2\) be? :D

Answer 15

2x...?

Answer 16

\[\large y'=\color{orangered}{(2x)}\cos x+x^2\color{royalblue}{(\cos x)'}\]

Good. So we've found the derivative of that first blue term.
The second term, there really isn't much of a process that we'll want to go through. You just need to memorize the derivative of cosine x.

Answer 17

Alrighty! So what's next?

Answer 18

What's the derivative of \(\large \cos x\) ?
Look it up in your notes somewhere if you have to! XD

Answer 19

Goodness...is it 0 or something? lol

Answer 20

noooooooooooo

:O

Answer 21

Oh nooo DX my notes are all messy.

Answer 22

http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions

Look at the chart in the upper right, it should be on that list.

Answer 23

Our `function` is \(\large \cos(x)\), what should the `derivative` be?

Answer 24

Ahaa so that's what my notes said. -sin(x)?

Answer 25

Yayyy you found it :D

\[\large y'=\color{orangered}{(2x)}\cos x+x^2\color{orangered}{(-\sin x)}\]

Answer 26

One final step, to clean it up might be to move the negative sign in front as subtraction. It will just look a little nicer.

Answer 27

\[\large y'=2x \cos x-x^2 \sin x\]

Answer 28

Awesome :D so that is it?

Answer 29

mhm -_-