Question

What is the derivative of f(x) = `sqrt(9-x^2)+4` considering the fact that the radius and tangent are perpendicular at the point of contact?

Answer 1

`(-x)/sqrt(9-x^2)`

Explanation:

f'(x) = `1/2 1/sqrt(9-x^2) (-2x) = (-x)/sqrt(9-x^2)`

The given function represents a circle of radius 3, centered at (0,4). This can be easily put into standard form as follows:

`y=f(x)= sqrt (9-x^2) +4` `->` `(y-4)^2 =9-x^2`

`x^2 +(y-4)^2= 3^2`

This equation represents a circle centered at (0,4) with radius 3.

In case of a circle, the radius joining the point of contact of any tangent is always perpendicular to the tangent,