Find `dy/dx`  by implicit differentiation. y sin(`x^2` ) = -x sin(`y^2` )

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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`ysin(x^2) = -xsin(y^2)`

`(ysin(x^2))' = (-xsin(y^2))'`

`(y')sin(x^2) + y(sin(x^2))' = (-x)' sin(y^2) + -x(sin(y^2))'`

`y'sin(x^2)+ y(2xcos(x^2)) = -sin(y^2) - x(2yy'cos(y^2))`

`y'sin(x^2) + 2xycos(x^2) = -sin(y^2) - 2xyy'cos(y^2)`

`y'sin(x^2) + 2xyy'cos(y^2) = -sin(y^2) - 2xycos(x^2)`

`y'( sin(x^2) + 2xycos(y^2)) = -sin(y^2) - 2xycos(x^2)`

`y' = (-sin(y^2) - 2xycos(x^2))/ (sin(x^2) + 2xycos(y^2))`

`y' = -(sin(y^2) + 2xycos(x^2))/ (sin(x^2)+2xycos(y^2))`

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beckden's profile pic

beckden | High School Teacher | (Level 1) Educator

Posted on

`ysin(x^2)=-xsin(y^2)`

`sin(x^2)dy+y(2xcos(x^2))dx=-sin(y^2)dx-x(2ycos(y^2))dy`

`(sin(x^2)+2xycos(y^2))dy=(-sin(y^2)-2xycos(x^2))dx`

` dy/dx=-(sin(y^2)+2xycos(x^2))/(sin(x^2)+2xycos(y^2))`

 

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