Calculus Help Please
Consider the implicit function
ln(xy)cos^1(x1)=x
a. Find dy/dx.
b. Find y when x = 1.
c. Find the equation of the tangent line in slopeintercept form when x = 1.
Show steps please! Thanks!

ln(xy)  cos^1(x1) = x
1/(xy) (y+xy') + 1/√(1(x1)^2) = 1
Now just solve for y'. You should get
y' = y(1  1/x  1/√(2xx^2))posted by 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.
ye^(1+pi/2) = e^(1+pi/2) (x1)
and convert that to slopeintercept.posted by Steve

ln x y = lnx + ln y
ln x + ln y cos^1(x1) = x
1/x + (1/y) dy/dx+(1/[1(x1)^2])^.5 = 1
1/x +(1/y)dy/dx+(1/(x^2+2x)^.5) = 1
(1/y) dy/dx = 1  1/x  [1/(x(2x))^.5
dy/dx = y { 1  1/x  [1/(x(2x))^.5 }
when x = 1
ln x + ln y cos^1(x1) = 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 + 2posted by Damon

oops  y' = y, so the tangent line is
y13 = 13(x1)
y = 13x+26
as shown at
http://www.wolframalpha.com/input/?i=plot+y+%3D+e^%28x%2Barccos%28x1%29%29%2Fx%2C+y%3D13x%2B26posted by Steve
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