solve dy/dx=x(2logx+1)/siny+ycosy .
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by cross multiplication {sin(y)+ycos(y)}dy=(2x*logx+x)dx
so derivative of y*sin(y)dy = derivative of [x^2logx]dx
so answer will be x^2logx
so derivative of y*sin(y)dy = derivative of [x^2logx]dx
so answer will be x^2logx
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