Question

Find the derivative of...
y=x^2sinx + 2cosx -2sinx
The answer is y'= x^2cosx but idk how to get that.

Answer 1

\[\Large y\quad=\quad x^2\sin x+2 \cos x- 2 \sin x\]
Taking the derivative of the last two terms is pretty straight forward right?
We need to apply the product rule to the first term.\[\Large \color{royalblue}{y'}\quad=\quad \color{royalblue}{(x^2)'}\sin x+x^2\color{royalblue}{(\sin x)'}+\color{royalblue}{(2\cos x)'}+\color{royalblue}{(-2\sin x)'}\]

Answer 2

How did you get x2(sinx)′?

Answer 3

\[\Large (x^2 \sin x)' \quad=\quad (x^2)'\sin x+ x^2 (\sin x)'\]Remember your product rule?\[\Large (fg)' =(f)'g+f(g)'\]

Answer 4

The blue parts are where you need to take a derivative.

Answer 5

i got \[2xsinx +x ^{2}cosx+2sinx-2cosx\] is that correct?

Answer 6

The third term:
(2cosx)' should give us -2sinx

The rest looks good though!

Answer 7

Idk what to do next now?:/

Answer 8

Hmm we're not going to get the answer you listed....
Are you sure there isn't more to this question? :d

Answer 9

yes i am sure.. hmm maybe its like this y=\[x ^{2}cosx-2sinx-(-sinx)\]