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Mathematics 6 Online
OpenStudy (anonymous):

Can anyone explain this for me? how did the equation (the left one) become like this after integral x^2 dy + 2yx dx = sin3x dx after integral yx^2 = -cos3x/3 + c

ganeshie8 (ganeshie8):

product rule

OpenStudy (anonymous):

reverse product rule

ganeshie8 (ganeshie8):

\((x^2y' + 2x y) = (x^2y)'\)

OpenStudy (anonymous):

if you divide both sides by dx , you'll get x^2 dy/dx + 2yx = sin3x on the left side, it's just d/dx (y x^2)

ganeshie8 (ganeshie8):

by FTC, \( \int (x^2y)' \) spits out \(x^2y\)

OpenStudy (anonymous):

thank you guys

ganeshie8 (ganeshie8):

np :)

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