# What is equation of tangent at curve f(x)=arctg x in point (1,f(1))?

sciencesolve | Certified Educator

You should use the following equation of tangent line to a given curve, at a given point `(x_0,y_0)` , such that:

`y - y_0 = f'(x_0)(x - x_0)`

Since the problem provides the tangency point `(1,f(1))` , you need to evaluate the value of the function at `x = 1` , such that:

`f(1) = arctan 1 = pi/4`

You need to evaluate the derivative of the function at `x = 1` , such that:

`f'(1) = lim_(x->1) (f(x) - f(1))/(x -1) = lim_(x->1) (arctan x - pi/4)/(x - 1) = (pi/4 - pi/4)/(1 - 1) = 0/0`

The indetermination case `0/0` requests for you to use l'Hospital's theorem, such that:

`lim_(x->1) (arctan x - pi/4)/(x - 1) = lim_(x->1) ((arctan x - pi/4)')/((x - 1)')`

`lim_(x->1) (arctan x - pi/4)/(x - 1) = lim_(x->1) (1/(1 + x^2))/1`

`lim_(x->1) (1/(1 + x^2)) = 1/(1+1) = 1/2`

Hence, evaluating the derivative of the function at `x = 1` , yields:

`f'(1) = 1/2`

Replacing `pi/4 ` for `f(1)` and `1/2` for `f'(1)` in equation of tangent line, yields:

`y - pi/4 = (1/2)(x - 1) => y = x/2 - 1/2 + pi/4`

Hence, evaluating the equation of tangent line to the curveĀ  `f(x) = arctan x` , at the point `(1,f(1))` , yields `y = x/2 - 1/2 + pi/4.`