# Use the derivitate function form: `f'(x) = lim_(h->0) (f(x+h)-f(x))/h` to find the tangent line to the graph of `f(x)=sqrt(x+1)` at `x=0` .

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Student Comments

aruv | Student

`f(x)=sqrt(x+1)`

`f(x+h)=sqrt(x+h+1)`

`f(x+h)-f(x)=sqrt(x+h+1)-sqrt(x+1)`

`But`

`lim_(h->0)(f(x+h)-f(x))/h=f'(x)`

`` Therefore

`f'(x)=lim_(h->o)(f(x+h)-f(x))/h=lim_(h->0)(sqrt(x+h+1)-sqrt(x+1))/h`

Multiply and divide by`sqrt(x+h+1)+sqrt(x+1)`

`f'(x)=lim_(h->0){(sqrt(x+h+1)-sqrt(x+1))(sqrt(x+h+1)+sqrt(x+1))}/(h(sqrt(x+h+1)+sqrt(x+1)))`

`=lim_(h->0)((x+h+1)-(x+1))/(h(sqrt(x+h+1)+sqrt(x+1)))`

`=1/(sqrt(x+0+1)+sqrt(x+1))`

`=1/(2sqrt(x+1))`

`` We need to draw tangent at point (0,f(0))

Thus

slope of the tangent at (0,f(0))

`m=f'(0)=1/(2sqrt(0+1))=1/2`

`f(0)=sqrt(0+1)=1`

`Thus`

Slope of the tangent at point (0,1) is 1/2.

Thus equation of the tangent is

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

2y-2=x

**Thus equation of the tangent at point (0,1) on the curve f(x)=sqrt(x+1) is**

** x-2y+2=0**