Find the equation of the line tangent to `f(x)=sqrt(x)/(6x-3)` when x=1:

`f(1)=sqrt(1)/(6-3)=1/3` so the point of tangency is `(1,1/3)` . To find the slope of the tangent line we evaluate `f'(1)` :

We can use the quotient rule to find `f'(x)` --

`f'(x)=((6x-3)(1/2)(x^(-1/2))-x^(1/2)(6))/(6x-3)^2`

It is unnecessary to simplify this; we only need to evaluate when x=1.

`f'(1)=(3*1/2)-6)/9=(-9/2)/9=-1/2`

So the point is `(1,1/3)` and the slope is `m=-1/2` so the equation of the tangent line is:

`y-1/3=-1/2(x-1)==>y=-1/2x+5/6`

-----------------------------------------------------------

The equation of the tangent line is `y=-1/2x+5/6`

----------------------------------------------------------

The graph of f(x) and the tangent line (in red):

` `