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
5 comments :
Thanks a lot for this post! I kept getting 1+tan^2(x) as the derivative of tan(x), and I thought I was having problems with the quotient rule, when really I was just simplifying it funny.
One could also use differentiation by parts;
d/dx(UV)=d/dx(U).V + d/dx(V).U
d/dx(tan(x))=sec^2(x)
As tan(x)=sin(x)/cos(x)
Take U=sin(x) , V=1/cos(x)=(cos(x))^-1
d/dx(U)=cos(x), d/dx(V)=sin(x)/cos^2(x)
therefore
d/dx(tan(x))=cos(x).1/cos(x)+sin^2(x)/cos^2(x)
=1+sin^2(x)/cos^2(x)
=[cos^2(x)+sin^2(x)]/cos^2(x)
as cos^2(x)+sin^2(x)=1
= 1/cos^2(x)
= (1/cos(x))^2
= (sec(x))^2
= sec^2(x)
The derivative of trigonometric functions is used in deriving solutions for real life mathematical functions and higher research work.
Derivative of Sec
Hey, Larry, this is a good post. I must say quotient rule is very useful to find the derivatives of trigonometric functions and also of algebraic functions. We can also find the derivatives of cosec x, sec x, cot x by using this rule otherwise substitution will also be another good option.
Very nice post but you didn't explain clearly proof of derivative of tan x by quotient, chain & first Principle rule
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