Question

What is the equation of the tangent line of `f(x)=sqrt(x+1)/sqrt(x+2)` at `x=4`?

Answer 1

Equation of tangent is `x-12sqrt30y+56=0`

Explanation:

Thr slope of a tangent to a curve `f(x)` at a point `x=x_0` is given by value of its derivative at `x=x_0` i.e. `f'(x_0)`.

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

Using quotient rule, `f'(x)=(sqrt(x+2)xx1/(2sqrt(x+1))-sqrt(x+1)xx1/(2sqrt(x+2)))/(x+2)`

and `f'(4)=(sqrt6/(2sqrt5)-sqrt5/(2sqrt6))/6`

= `(1/sqrt30)/12=1/(12sqrt30)`

Now at `x=4`, `f(x)=sqrt5/sqrt6`.

Hence, using point slope form, equation of desired tangent is

`y-sqrt5/sqrt6=1/(12sqrt30)(x-4)`

or `12sqrt30y-60=x-4`

or `x-12sqrt30y+56=0`

graph{(y-sqrt(x+1)/sqrt(x+2))(x-12sqrt30y+56)=0 [-0.29, 4.71, -0.53, 1.97]}