What is the equation of the line tangent to f(x)=sqrt(e^x-x) at x=1?

1 Answer
Nov 30, 2015

2y=sqrt(e-1)x +sqrt(e-1)
OR y=sqrt(e-1)x/2+sqrt(e-1)/2

Explanation:

f(x)= #sqrt(e^x-x)

First derivative at x=1 is the slope of the tangent to f(x) at x=1

f'(x)= (e^x-1)/(2sqrt(e^x-x))(using [chain rule](http://socratic.org/calculus/basic-differentiation-rules/chain-rule) )
f'(1)=(e-1)/(2sqrt(e-1))

f'(1) =(e-1)/2

slope =(e-1)/2 and points (1,f(1) ) ; f(1)=sqrt(e-1)

using equation of line in point slope form

y-y1=m(x-x1) ; m=slope=(e-1)/2 ;x1=1 and y1=sqrt(e-1)

y-sqrt(e-1)=(e-1)/2(x-1)

2y-2sqrt(e-1)=sqrt(e-1)x-sqrt(e-1)

2y=sqrt(e-1)x+sqrt(e-1)

y=sqrt(e-1)x/2+sqrt(e-1)/2