posted by Tenshi

Consider the implicit function

ln(xy)-cos^-1(x-1)=x

a. Find dy/dx.
b. Find y when x = 1.
c. Find the equation of the tangent line in slope-intercept form when x = 1.

1. Steve

ln(xy) - cos^-1(x-1) = x
1/(xy) (y+xy') + 1/√(1-(x-1)^2) = 1
Now just solve for y'. You should get

y' = y(1 - 1/x - 1/√(2x-x^2))

2. Steve

when x=1,
ln(y) - pi/2 = 1
y = e^(1+pi/2)

at x=1, y' = -y

so, now you have a point and a slope.

y-e^(1+pi/2) = -e^(1+pi/2) (x-1)
and convert that to slope-intercept.

3. Damon

ln x y = lnx + ln y

ln x + ln y -cos^-1(x-1) = x

1/x + (1/y) dy/dx+(1/[1-(x-1)^2])^.5 = 1

1/x +(1/y)dy/dx+(1/(-x^2+2x)^.5) = 1

(1/y) dy/dx = 1 - 1/x - [1/(x(2-x))^.5

dy/dx = y { 1 - 1/x - [1/(x(2-x))^.5 }

when x = 1
ln x + ln y -cos^-1(x-1) = x
0 + ln y - pi/2 = 1
ln y = 1 + pi/2
e^ln y = y = e^(1+pi/2)
y = 13.08 approx

slope = dy/dx call y = 13
= 13{1 -1/1 - [1/1] = 11

y = 11 x + b
13 = 11 + b
b = 2
y = 11 x + 2

4. Steve

oops --- y' = -y, so the tangent line is

y-13 = -13(x-1)
y = -13x+26

as shown at

http://www.wolframalpha.com/input/?i=plot+y+%3D+e^%28x%2Barccos%28x-1%29%29%2Fx%2C+y%3D-13x%2B26

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