Is there a way to prove the derivative of arctan(x) = 1 / (1 + x²) using the limit definition?

Question

Supposedly any derivative can be taken using the limit definition of a derivative. Here's one that looks tricky. Good luck.

Answer 1

Here is the approach, some details remain for You.
Start with the these very well known:
1) lim[z→0] sin z / z = 1;
2) tan(α - β) = (tan α - tan β)/(1 + tan α * tan β).
From 1) easily follows:
lim[z→0] tan z / z = 1, hence lim[z→0] arctan(z) / z = 1.
From 2) and some other assumptions:
arctan(A) - arctan(B) = arctan((A - B)/(1 + AB))

Now we are ready for the main thing "ab ovo":
(arctan(x))' = lim[h→x](arctan(x) - arctan(h))/(x - h) =
= lim[h→x] arctan((x - h)/(1 + hx))/(x - h) =
= lim[h→x] arctan((x - h)/(1 + hx))/((x - h)/(1 + hx)) *
* lim[h→x] 1/(1 + hx) = 1 * lim[h→x] 1/(1 + hx) =
= 1/(1 + x²) as required.

Answer 2

Arctan Definition

Answer 3

f(x) = (x+1)/(x-1) f(x+h) = (x+h+1)/(x+h-1) f(x+h) - f(x) = (x+h+1)/(x+h-1) - (x+1)/(x-1) = ( (x-1)(x+h+1) - (x+h-1)(x+1) ) / (x+h-1)(x-1) = ( x² + hx + x - x - h - 1 - x² - x - hx - h + x + 1) / (x+h-1)(x-1) = -2h / (x + h - 1)(x - 1) ( f(x+h) - f(x) ) / h = -2 / (x + h - 1)(x - 1) The limit as h approaches 0 is: -2 / (x-1)(x-1) = -2/(x-1)²

Answer 4

yes but it requires a little bit of changing around.

let y = arctan x or tan y =x.
Now we can take the derivative of the latter version of the equation with respect to x

(d/dx)[x] = (d/dx)[tan y]

The left side will be 1 but for the right side we will have to use implicit differentiation. So

1 = (d/dx)[tan y] = (d/dy)[tan y]*(dy/dx).

Here's is where you can use the limit definition of the derivative to show to that (d/dy)[tan y] = sec^2(y). From this

1 = sec^2(y)*(dy/dx) or
(dy/dx) = 1/[sec^2(y)]
= 1/[1+ tan^2(y)]
= 1/(1+x^2), since we had tan y = x