How do you calculate #y^2x = 2sin(xy^3) #For implicit differentiation? Calculus Basic Differentiation Rules Implicit Differentiation 1 Answer Narad T. · Steve M May 17, 2018 The answer is #=(y(2ycos(xy^3-1)))/(2x(1-3ysin(xy^3)))# Explanation: Let #f(x,y)=y^2x-2sin(xy^3)# Then, #(delf)/(delx)=y^2-2y^3cos(xy^3)=y^2(1-2ycos(xy^3))# #(delf)/(dely)=2yx-6xy^2sin(xy^3)=2yx(1-3ysin(xy^3))# Therefore, #dy/dx=-((delf)/(delx))/((delf)/(dely))=-(y^2(1-2ycos(xy^3)))/(2yx(1-3ysin(xy^3)))# #=(y(2ycos(xy^3)-1))/(2x(1-3ysin(xy^3)))# Answer link Related questions What is implicit differentiation? How do you find the derivative using implicit differentiation? How do you find the second derivative by implicit differentiation? How do you find #y''# by implicit differentiation of #x^3+y^3=1# ? How does implicit differentiation work? How do you use implicit differentiation to find #(d^2y)/dx^2# of #x^3+y^3=1# ? How do you Use implicit differentiation to find the equation of the tangent line to the curve... How do you use implicit differentiation to find #y'# for #sin(xy) = 1#? How do you find the second derivative by implicit differentiation on #x^3y^3=8# ? What is the derivative of #x=y^2#? See all questions in Implicit Differentiation Impact of this question 427 views around the world You can reuse this answer Creative Commons License