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

Question

Scarlett Allison · Accepted Answer

The slope of the tangent is #0# (and #C# is not "arbitrary", it is #-3#).

#(x-y)(x+y)-xy+y= C # Instead of differentiating immediately, let's rewrite the first product. #x^2-y^2-xy+y=C#. This form is (I think) simpler to differentiate. #2x-2y dy/dx - y - x dy/dx + dy/dx = 0# At the point #(1,2)#, we get #2(1) - 2(2) dy/dx - 2 - (1) dy/dx + dy/dx = 0# Solve for #dy/dx# to see that #dy/dx = 0#

Cameron Alexander · Answer

undefined To find the slope of the tangent line at the point (1,2), you first need to find the derivative of the given function with respect to x. Then, evaluate the derivative at x = 1 to find the slope. The derivative of the function (x-y)(x+y)-xy+y=C with respect to x is found by applying the chain rule and product rule. After finding the derivative, substitute x = 1 and y = 2 into the derivative expression to get the slope of the tangent line at (1,2).

William Abdalla · Answer

undefined To find the slope of the tangent line of the curve at a given point, we first need to find the derivative of the curve with respect to x. Then, we evaluate this derivative at the given point to find the slope of the tangent line.

Given the curve equation (x-y)(x+y)-xy+y=C, we differentiate it with respect to x:

d/dx [(x-y)(x+y)-xy+y] = 0

Then, we evaluate this derivative at the point (1,2):

(x-y)(1+y) + (x+y)(1-y) - y - x(dy/dx) = 0

Substitute x=1 and y=2 into this equation and solve for dy/dx, which represents the slope of the tangent line at (1,2).