Asked by John
How would you find the derivative of this implicit function?
1+x=sin(xy^2)
1+x=sin(xy^2)
Answers
Answered by
Yu
Use Chain Rule
1=cos(xy^2)(y^2+2xyy')
1=cos(xy^2)y^2+(2xy)cos(xy^2)(y')
[1-cos(xy^2)y^2]/[(2xy)cos(xy^2)]=y'
1=cos(xy^2)(y^2+2xyy')
1=cos(xy^2)y^2+(2xy)cos(xy^2)(y')
[1-cos(xy^2)y^2]/[(2xy)cos(xy^2)]=y'
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