Answer:
B) L(x) = -x + 2
Step-by-step explanation:
To find the tangent at (1,1), we differentiate x² - ln(x/y) = 1 with resect to x.
So, x² - ln(x/y) = 1
x² - (lnx - lny) = 1
x² - lnx + lny = 1
differentiating both sides with respect to x, we have
d{x² - lnx + lny]/dx = d1/dx
dx²/dx - dlnx/dx + dlny]/dx = d1/dx
2x - 1/x + 1/y(dy/dx) = 0
1/y(dy/dx) = 1/x - 2x
dy/dx = y(1/x - 2x)
dy/dx evaluated at (1,1) is
dy/dx = y(1/x - 2x)
= (1)(1/1 - 2(1))
= 1 - 2
= -1
This is the gradient of the tangent line.
To find the equation of the tangent line, we use (y - y')/(x - x') = m where y' = 1, x' = 1 and m = gradient of line = dy/dx = - 1
(y - y')/(x - x') = m
(y - y')/(x - x') = dy/dx
(y - 1)/(x - 1) = -1
cross-multiplying,
y - 1 = -1(x - 1)
expanding the bracket, we have
y - 1 = -x + 1
adding + 1 to both sides, we have
y -1 + 1 = -x + 1 + 1
y = -x + 2
So, L(x) = -x + 2
