Please Help! Medal Will Be Rewarded! Find y’ for y=x sin x^2
y'=x*2 sin x cos x+sinx^2
Thank you so much!!!!
Is it \[ x \sin(x^2) \] or \[ x (\sin (x))^2 \]
correction I solved \[y=x(sinx)^{2} \]
Its x sin (x^2)
So the answer is like the one obtained with wolframalpha
that is what i am wondering - is it the one from wolfram?
\[if y=x \sin x ^{2},y'=x \left( \cos x ^{2}*2x \right)+\sin x ^{2}\]
Are you positive? Just checking as http://www5a.wolframalpha.com/Calculate/MSP/MSP35281b7b6cc0d3ic7gf9000025e15d6ah6e9egai?MSPStoreType=image/gif&s=34&w=239.&h=36.
@surjithayer second solution is correct \[ y'=\sin \left(x^2\right)+2 x^2 \cos \left(x^2\right) \]
here after simplification \[y'=2x ^{2}\cos x ^{2}+\sin x ^{2}\]
Okay, thank you @surjithayer !!! I have more questions to post if you guys want more medals!
I am here to help not for medals.
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