How do you implicitly differentiate #3y + y^4/x^2 = 2#?
Thanks for your help!
Thanks for your help!
1 Answer
Dec 8, 2016
For problems like these, we must remember that we're differentiating with respect to x.
#3 + (4y^3x^2(dy/dx) - 2x(y^4))/(x^2)^2 = 0#
#3(dy/dx) + (4y^3x^2(dy/dx)- 2xy^4)/x^4 = 0#
#(3x^4(dy/dx) + 4y^3x^2(dy/dx) - 2xy^4)/x^4 =0#
#3x^4(dy/dx) + 4y^3x^2(dy/dx) - 2xy^4 =0#
#3x^4(dy/dx) + 4y^3x^2(dy/dx) = 2xy^4#
#dy/dx(3x^4 + 4y^3x^2) = 2xy^4#
#dy/dx = (2xy^4)/(3x^4 + 4y^3x^2)#
#dy/dx = (2xy^4)/(x^2(3x^2 + 4y^3))#
#dy/dx = (2y^4)/(3x^3 + 4xy^3)#
Hopefully this helps!
