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

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Answer 1 · 2016-12-19T12:19:04

#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]}