What is the slope of the tangent line of #1/x^3-y^2-y = C #, where C is an arbitrary constant, at #(-5,0)#?

1 Answer
Nov 22, 2015

The slope is #(-3)/625#

Explanation:

The constant #C#
If we know that #(-5,0)# is a point on the graph, then #C = -1/125#.
So #C# is determined by the values of #x# and #y#. It is not arbitrary. (It is not independent of the values of the variables.)

Slope of Tangent
Regardless of that, we can find the slope of the tangent line by implicit differentiation.

#d/dx(1/x^3-y^2-y)=d/dx(C) #

#-3/x^4-2ydy/dx=dy/dx = 0#

#-3/x^4 = (2y+1) dy/dx#

The slope at #(-5,0)#, is therefore

#dy/dx = -3/625#