How do you find the slope of the line tangent to #y+e^y=1+lnx# at #(1,0)#?

Question

Christopher Agresta · Accepted Answer

#y=x/2-1/2#

The point of contact of the tangent is #P (1, 0)#.

(y+e^y)'=(1+e^y)y'=(1=ln x)'=1/x#

At P (1, 0), the slope of the tangent

#m = y'at P= 1/2.

So, the equation of the tangent is

#y-0=1/2(x-1)#

P is on the x-axis.

graph{y+e^y-1-ln x=0 [-10, 10, -5, 5]}

Christopher Agresta · Answer

undefined To find the slope of the line tangent to the curve y + e^y = 1 + ln(x) at the point (1,0), we can use the concept of implicit differentiation.

First, differentiate both sides of the equation with respect to x:

d/dx(y) + d/dx(e^y) = d/dx(1 + ln(x))

Since y is a function of x, we can use the chain rule to differentiate y with respect to x:

dy/dx + d/dx(e^y) = 0 + 1/x

Next, we need to find the derivative of e^y with respect to x. Using the chain rule again, we have:

d/dx(e^y) = d/dy(e^y) * dy/dx

The derivative of e^y with respect to y is simply e^y, and dy/dx is the derivative of y with respect to x. Substituting these values back into the equation, we get:

dy/dx + e^y * dy/dx = 1/x

Now, we can solve for dy/dx, which represents the slope of the tangent line at the point (1,0). Rearranging the equation, we have:

dy/dx * (1 + e^y) = 1/x

dy/dx = 1/x / (1 + e^y)

To find the slope at the point (1,0), substitute x = 1 and y = 0 into the equation:

dy/dx = 1/1 / (1 + e^0)

dy/dx = 1 / (1 + 1)

dy/dx = 1/2

Therefore, the slope of the line tangent to y + e^y = 1 + ln(x) at the point (1,0) is 1/2.