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.
Show steps please! Thanks!
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))
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.
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
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
To find the derivative of the implicit function ln(xy) - cos^(-1)(x-1) = x with respect to x, we can use the chain rule and implicit differentiation.
a. Find dy/dx:
Step 1: Take the derivative of both sides of the equation with respect to x.
d/dx[ln(xy) - cos^(-1)(x-1)] = d/dx[x]
Step 2: Apply the chain rule:
1/(xy) * (y + xy') - d/dx[cos^(-1)(x-1)] = 1
Step 3: Simplify and solve for dy/dx:
(y + xy')/(xy) = d/dx[cos^(-1)(x-1)]
Step 4: Solve for dy/dx:
(y + xy')/(xy) = -1/sqrt(1 - (x-1)^2)
Step 5: Multiply both sides by (xy):
y + xy' = -xy/sqrt(1 - (x-1)^2)
Step 6: Rearrange the equation to isolate dy/dx:
xy' = -xy/sqrt(1 - (x-1)^2) - y
Step 7: Divide both sides by x:
y' = -y/x - y/(x * sqrt(1 - (x-1)^2))
Therefore, dy/dx = -y/x - y/(x * sqrt(1 - (x-1)^2))
b. To find y when x = 1, substitute x = 1 into the implicit function and solve for y:
ln(1*y) - cos^(-1)(1-1) = 1
ln(y) - cos^(-1)(0) = 1
ln(y) = 1
Exponentiate both sides to eliminate the natural logarithm:
y = e^1
y = e
Therefore, when x = 1, y = e.
c. To find the equation of the tangent line in slope-intercept form when x = 1, we need to find the slope and the y-intercept.
Slope:
Evaluate dy/dx at x = 1:
dy/dx = -y/1 - y/(1 * sqrt(1 - (1-1)^2))
= -y - y/0 (since the denominator is zero, we need to evaluate the limit)
= -∞
The slope of the tangent line is -∞.
Y-intercept:
Substitute x = 1 and y = e into the point-slope form equation:
y - y1 = m(x - x1)
y - e = -∞(x - 1)
Simplifying this equation, the y-intercept form is:
y = -∞x + ∞e
Therefore, the equation of the tangent line when x = 1 is y = -∞x + ∞e.