calculus question

Question

Find an equation of the tangent line to the graph of the function at the given point.

f(x) = e^−x ln(x),    (1, 0)

Answer 1

Tangent lines are lines that just touch the point you are talking about that match the slope of the curve at that point. So the first step is to find the slope of the point. Then you can use that slope to fit the equation for a line (y - yp) = m (x - xp). Once you plug in all the numbers, you get the tangent line.

So for this problem, you have the curve, and you need to find the slope at (1,0). In order to do that you need to find the first derivative of the equation. f(x) = e^-x * ln(x), so using the product rule you get that f'(x) = e^-x (1/x) - e^-x * ln(x). Solving for f'(1), you get that f'(1) = e^-1 = 1/e. This is your slope. Then you can plug into the line equation (y-0) = (1/e) (x - 1). This is the tangent line and if needed you can simplify the equation more and put it into the form y = mx + b if needed.

Answer 2

y' = exp(-x)/x - exp(-x)lnx = f'(x)

m = f'(1) = 1/e

B = y - mx

= 0 - (1/e)(1)

= -1/e

y = (1/e)x - 1/e