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OpenStudy (anonymous):

what is the derivative of sin(xy^2)

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OpenStudy (aum):

Assuming derivative with respect to x: \[ \frac{d}{dx}\sin(xy^2) = \cos(xy^2) * \frac{d}{dx}(xy^2) = \cos(xy^2) * \{x * 2y * \frac{dy}{dx} + 1 * y^2 \} = \\ 2xy\cos(xy^2)\frac{dy}{dx} + y^2\cos(xy^2) \]

OpenStudy (anonymous):

Okay, thank you, I understand most of it, but I do have another question. How did you get the last {cos(xy^2)} that is with the y^2?

OpenStudy (aum):

( sin(xy^2) )' = cos(xy^2) * (xy^2)' Apply the product rule to find (xy^2)': x * 2yy' + 1*y^2 ( sin(xy^2) )' = cos(xy^2) * (xy^2)' = cos(xy^2) * { x * 2yy' + 1*y^2 } multiply it out: cos(xy^2) * x * 2yy' + cos(xy^2) * y^2 = 2xycos(xy^2)y' + y^2cos(xy^2)

OpenStudy (anonymous):

I can't thank you enough, you are the best, thanks

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