OpenStudy (anonymous):

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

4 years ago
zepdrix (zepdrix):

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

4 years ago
OpenStudy (anonymous):

How did you get x2(sinx)′?

4 years ago
zepdrix (zepdrix):

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

4 years ago
zepdrix (zepdrix):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
zepdrix (zepdrix):

The third term: (2cosx)' should give us -2sinx The rest looks good though!

4 years ago
OpenStudy (anonymous):

Idk what to do next now?:/

4 years ago
zepdrix (zepdrix):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
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