Find an equation of the tangent line to the graph of the function at the given point.
1 + ln xy = e^(x-y), (1,1)
To find the equation of the tangent line to the graph of a function at a given point, you need to determine the slope of the tangent line and use the point-slope form of a linear equation.
To find the slope of the tangent line, you can use the derivative of the function. So, let's start by finding the derivative of the function 1 + ln(xy) = e^(x-y).
First, let's rewrite the equation in a more convenient form:
ln(xy) = e^(x-y) - 1
To find the derivative, you need to apply the chain rule because you have a composition of functions. The derivative of ln(xy) with respect to x can be found as follows:
d/dx[ln(xy)] = (1/xy)(dy/dx)
The derivative of e^(x-y) - 1 with respect to x can be found using the chain rule:
d/dx[e^(x-y) - 1] = (e^(x-y) - 1)(1) - (e^(x-y))(d/dx(x-y))
Simplifying the derivative, we have:
(1/xy)(dy/dx) = (e^(x-y) - 1) - (e^(x-y))(1-0)
Now, substitute the given point (1, 1) into the equation to find the value of dy/dx at that point:
(1/1)(dy/dx) = (e^(1-1) - 1) - (e^(1-1))(1-0)
dy/dx = e^0 - 1 - e^0
dy/dx = 1 - 1 - 1
dy/dx = -1
The slope of the tangent line to the graph of the function at the point (1, 1) is -1.
Now, we can use the slope-intercept form of a linear equation, y = mx + b, to find the equation of the tangent line. Substituting the values into the equation, we have:
1 = -1(1) + b
1 = -1 + b
b = 2
Therefore, the equation of the tangent line is y = -x + 2.