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

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

• Edwards & Penny, Calculus and Analytic Geometry