## Tuesday, November 27, 2007

### Derivative of tan x

Theorem: Derivative of tan x = sec2x

Proof:

(1) d(sin x)/dx = cos x [See Theorem 1, here]

(2) d(cos x)/dx = -sin x [See Theorem 2, here]

(3) Now, tan x = (sin x)/(cos x)

(4) Let f(x) = sin x, let g(x) = cos x

(5) If we assume that x ≠ ± π/2 and x is between π/2 and -π/2, then we can see that f(x), g(x) is differentiable for all values of x and further that g(x) ≠ 0. [See here for review of cos, sin if needed]

(6) From step #5, we can now apply the Quotient Rule (see Lemma 6, here) to get:

d(tan x)/dx = [f'(x)g(x) - f(x)g'(x)]/[g(x)]2 = [cos x*cos x - sin x*(- sin x)]/[cos x]2 = [cos2 x + sin2 x]/(cos2 x)

(7) Now, using sin2x + cos2x = 1 [See Corollary 2, here], we have:

d(tan x)/dx = 1/(cos2 x)

(8) Since sec x = 1/cos x, we now have:

d(tan x)/dx = sec2 x

QED

## Monday, November 26, 2007

### Inverse Tangent

Definition 1: Inverse Tangent

y = tan-1x if and only if tan y = x where -π/2 is less than y is less than π/2.

Lemma: sec2x = 1 + tan2 x

Proof:

(1) By definition sec x = 1/cos x

(2) So, sec2x = 1/(cos2 x)

(3) sin2x + cos2x = 1 [See Corollary 2, here]

(4) 1/(cos2 x) = (cos2x + sin2x)/(cos2) =

= cos2x/(cos2x) + (sin2x)/(cos2) = 1 + tan2x.

QED

Theorem: D tan-1 x = 1/(1 + x2)

Proof:

(1) Let y = tan-1 x

(2) Then tan y = x [See Definition 1 above]

(3) d(tan y)/dx = d(x)/dx = 1

(4) d(tan y)/dx = (sec2y)dy/dx = 1 [See Theorem 1, here]

(5) dy/dx = 1/sec2y

(6) Using Lemma 1 above, we have:

dy/dx =1/(1 + tan2y)

(6) Using step #2, this gives us:

dy/dx = 1/(1 + tan2y) = 1/(1 + x2) [Since x = tan y]

QED

References