What is the slope of the tangent line of #x/y^3-xy= C #, where C is an arbitrary constant, at #(1,2)#?

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Answer 1 · 2018-07-27T12:40:39

Slope of the tangent line at #(1,2)# is #-1.58#

# x/y^3-x y = C; (1,2)# or

# x*y^(-3)-x y = C; # , differentiating we get,

# y^(-3)+ x (-3 y^-4)y^'- y-x y^' = 0; # or

#y^'(-(3 x)/y^4 - x)= y- 1/y^3 #

#:. y^'= (y- 1/y^3)/(-((3 x)/y^4 + x) #; Putting# x=1 ,y=2#

we get ,#y^'= (2- 1/2^3)/(-((3*1)/2^4 + 1)) or y^'~~ -1.58#

Slope of the tangent line at #(1,2)# is # m ~~ -1.58# [Ans]