Question

What is the equation of the tangent line of `f(x)=sqrt(x+1)-sqrt(x+2)` at `x=2`?

Answer 1

`4ysqrt3=x(2-sqrt3)+8-6sqrt3`

Explanation:

`f(x)=sqrt(x+1)-sqrt(x+2)`
`f'(x)=1/(2sqrt(x+1))-1/(2sqrt(x+2)`
`f'(x)=(1/2)(sqrt(x+2)-sqrt(x+1))/(sqrt(x+1)sqrt(x+2))`

At `(2,sqrt3-2)`

Sub x=2 into `f'(x)`

`f'(2)=(1/2)(sqrt(2+2)-sqrt(2+1))/(sqrt(2+1)sqrt(2+2))`
`f'(2)=(2-sqrt3)/(4sqrt3)`

Equation of the tangent:

`(y-sqrt3+2)=(2-sqrt3)/(4sqrt3)(x-2)`
`4ysqrt3-12+8sqrt3=2x-4-xsqrt3+2sqrt3`
`4ysqrt3=x(2-sqrt3)+8-6sqrt3`