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Mathematics 9 Online
OpenStudy (anonymous):

Please Help! Medal Will Be Rewarded! Find y’ for y=x sin x^2

OpenStudy (anonymous):

y'=x*2 sin x cos x+sinx^2

OpenStudy (anonymous):

Thank you so much!!!!

OpenStudy (anonymous):

Is it \[ x \sin(x^2) \] or \[ x (\sin (x))^2 \]

OpenStudy (anonymous):

correction I solved \[y=x(sinx)^{2} \]

OpenStudy (anonymous):

Its x sin (x^2)

OpenStudy (anonymous):

So the answer is like the one obtained with wolframalpha

OpenStudy (anonymous):

that is what i am wondering - is it the one from wolfram?

OpenStudy (anonymous):

\[if y=x \sin x ^{2},y'=x \left( \cos x ^{2}*2x \right)+\sin x ^{2}\]

OpenStudy (anonymous):

@surjithayer second solution is correct \[ y'=\sin \left(x^2\right)+2 x^2 \cos \left(x^2\right) \]

OpenStudy (anonymous):

here after simplification \[y'=2x ^{2}\cos x ^{2}+\sin x ^{2}\]

OpenStudy (anonymous):

Okay, thank you @surjithayer !!! I have more questions to post if you guys want more medals!

OpenStudy (anonymous):

I am here to help not for medals.

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