
Steve S. answered 03/10/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
x^6 y^2 - 4 = 0
Later we may need: y^2 = 4/x^6; y = ± 2/(x^3)
Differentiate implicitly:
(x^6 y^2)' - 0 = 0
Use Product Rule:
(x^6)'(y^2) + (x^6)(y^2)' = 0
(6x^5)(y^2) + (x^6)(2y)(y’) = 0
2(x^5)(y)(3y + xy’) = 0
3y + xy’ = 0
Later we may need: y’ = –3y/x.
Differentiate implicitly again:
3y’ + y’ + xy’’ = 0
4y’ + xy’’ = 0
Substitute y’ = –3y/x:
4(–3y/x) + xy’’ = 0
-12y/x + xy’’ = 0
xy’’ = 12y/x
y’’ = 12y/(x^2)
Substitute y = ± 2/(x^3):
y’’ = 12(± 2/(x^3))/(x^2)
y’’ = ± 24/(x^5)
Later we may need: y^2 = 4/x^6; y = ± 2/(x^3)
Differentiate implicitly:
(x^6 y^2)' - 0 = 0
Use Product Rule:
(x^6)'(y^2) + (x^6)(y^2)' = 0
(6x^5)(y^2) + (x^6)(2y)(y’) = 0
2(x^5)(y)(3y + xy’) = 0
3y + xy’ = 0
Later we may need: y’ = –3y/x.
Differentiate implicitly again:
3y’ + y’ + xy’’ = 0
4y’ + xy’’ = 0
Substitute y’ = –3y/x:
4(–3y/x) + xy’’ = 0
-12y/x + xy’’ = 0
xy’’ = 12y/x
y’’ = 12y/(x^2)
Substitute y = ± 2/(x^3):
y’’ = 12(± 2/(x^3))/(x^2)
y’’ = ± 24/(x^5)