6. Given the implicit curve x^2-ln(x/y)=1 , Write the linear approximation () of the tangent line at (1,1) A) L(x)=-x-2 B) L(x)= -x+2 C) L(x)= -x-1 D) L(x)= -x+1

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

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