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)#?

Question

Emilia Aguilar · Accepted Answer

#(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#

Ezekiel Alexander · Answer

undefined To find the slope of the tangent line at a given point on a curve, you need to first differentiate the equation of the curve with respect to $ x $ to find the derivative, and then evaluate the derivative at the given point to find the slope of the tangent line.

Given the equation $ x^3 - \frac{x+y}{x-y} = C $, we differentiate both sides with respect to $ x $ to find the derivative.

$$ \frac{d}{dx}\left(x^3 - \frac{x+y}{x-y}\right) = \frac{d}{dx}(C) $$

$$ 3x^2 - \frac{(x-y) \frac{d}{dx}(x+y) - (x+y) \frac{d}{dx}(x-y)}{(x-y)^2} = 0 $$

$$ 3x^2 - \frac{(x-y)(1+1) - (x+y)(1-1)}{(x-y)^2} = 0 $$

$$ 3x^2 - \frac{2x - 2y}{(x-y)^2} = 0 $$

$$ 3x^2(x-y)^2 - (2x - 2y) = 0 $$

$$ 3x^2(x-y)^2 = 2x - 2y $$

$$ (x-y)^2 = \frac{2x - 2y}{3x^2} $$

$$ x - y = \pm \sqrt{\frac{2x - 2y}{3x^2}} $$

Now, to find the slope of the tangent line at the point (1,4), substitute $ x = 1 $ and $ y = 4 $ into the equation we've derived and find the slope.