Question

Find the equation of a line tangent to the curve xy = (suqareroot^xy -x) + 1 at the point (1, 2).

A. y= 2-x
B. y = 5-3x
C. y = 5.667 - 2.667x
D. y= 2.667 - 1.333x
E. y = 2.667 - 1.667x

Answer 1

clarify what is meant by " (suqareroot^xy -x) "
is it √(xy) - x
is it (√x)y - x

√<super>xy</super> is meaningless

Answer 2

that last part was supposed to look like this:

√^xy is meaningless

Answer 3

or in better format
(√)^(xy)

Answer 4

xy=√xy-x +1

Answer 5

I assume you meant
xy = √(xy-x) + 1
because (1,2) actually satisfies the equation.
So, now we have
y + xy' = 1/(2√(xy-x)) * (y+xy')
y+xy' = 1 - 1/(2(xy-1))
y' = (1-2y+2xy^2)/(3x-2x^2y)
at (1,2) then, y' = -5
so the tangent line is
y-2 = -5(x-1)
y = 7-5x

Hmmm. Better double-check my math, and my interpretation of the equation.

Answer 6

B. y = 5-3x