How do you find an equation of the tangent line to the curve #y=e^x/x# at the point (1,e)?

Question

Christopher Agresta · Accepted Answer

Find the derivative and plug in our x-coordinate to find the slope of the tangent line. Then use that slope for a point-slope formula.

#y'=(e^x-xe^x)/x^2# #y'(1)=0=m# This is our slope; a horizontal line. #y-y_1=m(x-x_1)# Since our slope(#m#) is 0, our line will just be our y-coordinate: #e# Therefore: The equation of the tangent line is: #y=e#

Elijah Abram · Answer

undefined To find the equation of the tangent line to the curve y=e^x/x at the point (1,e), we need to find the slope of the tangent line and the coordinates of the point.

First, let's find the slope of the tangent line. We can do this by taking the derivative of the function y=e^x/x. Using the quotient rule, the derivative is given by:

dy/dx = (x(e^x) - e^x)/x^2

Next, substitute x=1 into the derivative to find the slope at the point (1,e):

dy/dx = (1(e^1) - e^1)/1^2 = (e - e)/1 = 0

The slope of the tangent line at the point (1,e) is 0.

Now, let's find the coordinates of the point. We already know that the x-coordinate is 1. To find the y-coordinate, substitute x=1 into the original function:

y = e^1/1 = e

So, the coordinates of the point are (1,e).

Now that we have the slope (0) and the coordinates of the point (1,e), we can use the point-slope form of a linear equation to find the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values, we get:

y - e = 0(x - 1)

Simplifying, we have:

y - e = 0

Therefore, the equation of the tangent line to the curve y=e^x/x at the point (1,e) is y = e.

Axel Alvarez · Answer

undefined To find the equation of the tangent line to the curve $y = \frac{e^x}{x}$ at the point $(1,e)$, follow these steps:

1. Find the derivative of the function $y = \frac{e^x}{x}$ using the quotient rule.
2. Evaluate the derivative at $x = 1$ to find the slope of the tangent line.
3. Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the given point, to write the equation of the tangent line.