Find an equation of the tangent line to the graph of te function y = 1+ln xex atthe point where x = 1

To find the equation of the tangent line to the graph of the function y = 1 + ln(xe^x) at the point where x = 1, we need to find the derivative of the function first.

Step 1: Differentiate the function The derivative of y = 1 + ln(xe^x) is y' = (1/x) + e^x.

Step 2: Evaluate the derivative at x = 1 Substitute x = 1 into the derivative to find the slope of the tangent line at x = 1. y'(1) = (1/1) + e^1 = 1 + e.

Step 3: Find the y-coordinate of the point where x = 1 Substitute x = 1 into the original function to find the y-coordinate. y(1) = 1 + ln(e) = 1 + 1 = 2.

Step 4: Use the point-slope form of the equation of a line to find the equation of the tangent line The point-slope form of the equation of a line is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Substitute m = 1 + e, x1 = 1, and y1 = 2 into the equation to find the equation of the tangent line. y - 2 = (1 + e)(x - 1).

So, the equation of the tangent line to the graph of the function y = 1 + ln(xe^x) at the point where x = 1 is y - 2 = (1 + e)(x - 1).