Determine the equation of the tangent line to (lny)^2 - e^xy at (0,e)?

Answer 1

What we want to find first is the slope of the equation at (0,e)

since we don't have an expression of y in terms of x we use implicit differentiation.

I have to assume here that (lny)^2 - e^xy = 0

since you don't have an equation at all unless it is equal to something.

d[(lny^2] - d[e^(xy)] = 0

d[f(y)^2] = 2f(y) df(y) = 2f(y)f'(y) dy
f(y) = ln(y) so f[ln(y)^2] = 2ln(y) * 1/y * dy

let xy=u
d[f(u)] = f'(u)d(u) = f'(xy)(ydx + xdy)
f(xy) = e^(xy) so d[e^(xy)] = e^xy(ydx + xdy)

so we have

2ln(y)/y dy - e^(xy)(ydx + xdy) = 0



grouping terms...

2ln(y)/y dy - xe^(xy)dy = ye^(xy)dx
2ln(y) dy - xye^(xy)dy = y^2e^(xy)dx
dy[2ln(y) - xye^(xy)] = y^2e^(xy)dx

dy/dx = y^2e^(xy)/[2ln(y) - xye^(xy)]
dy/dx = y^2/[2ln(y)e^(-xy) - xy]

So the slope at (0,e) is determined by substituting x=0 and y=e into dy/dx and you get

dy/dx = e^2/[2ln(e)] = e^2/2 at (0,e)

A line with this slope that passes through this point is on the tangent line.

Now that we have the slope of the line we then determine the equation of the line. The equation of a line with slope m that passeses through a point (x1,y1) is

y - y1 = m(x - x1)
y - e = e^2/2 (x - 0)

y = e^2/2 x+ e

And that's your line of course that is using the assumption that your original equation was equal to 0. If it is equal to something else you can just plug i in and follow the steps above.