What is the slope of the tangent line of #x^3-(x+y)/(x-y)= C #, where C is an arbitrary constant, at #(1,4)#?

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Answer 1 · 2018-07-18T17:23:35

#(dy/dx)_{(1, 4)}=35/2#

Since, the point #(1, 4)# lies on the curve: #x^3-\frac{x+y}{x-y}=C# hence it will satisfy the equation of curve as follows

#1^3-\frac{1+4}{1-4}=C#

#C=8/3#

hence, substituting above value, the equation of given curve is

#x^3-\frac{x+y}{x-y}=8/3#

#y=\frac{3x^4-11x}{3x^3-5}#

differentiating above equation w.r.t. #x# using division rule, we get the slope

#dy/dx=d/dx(\frac{3x^4-11x}{3x^3-5})#

#=\frac{(3x^3-5)d/dx(3x^4-11x)-(3x^4-11x)d/dx(3x^3-5)}{(3x^3-5)^2}#

#=\frac{(3x^3-5)(12x^3-11)-(3x^4-11x)(9x^2)}{(3x^3-5)^2}#

Hence, substituting #x=1# in above equation, the slope #dy/dx# of tangent at #(1, 4)# is

#(dy/dx)_{(1, 4)}=\frac{(3(1)^3-5)(12(1)^3-11)-(3(1)^4-11\cdot 1)(9(1)^2)}{(3(1)^3-5)^2}#

#=35/2#