Derivative of Tan x

Text-only Preview

Derivative of Tan x
Derivative of Tan x
The derivative of tan(X)
Since tan(X)=sin(X)/cos(X), we have sin(X) as the function u(X) and cos(X) as the
function v(X). Putting these into the formula
d[uv]/dX=(vdu/dX - udv/dX)/v2
we get
d[tan(X)]/dX = (cos(X)cos(X) + sin(X)sin(X))/cos2(X)
But on the top we have sin2(X)+ cos2(X), which is always 1. So our result simplifies
to d[tan(X)]/dX = 1/cos2(X).
But that is sec2(X), since sec(X)=1/cos(X).
So the derivative of tan(X) is sec2(X).
KnowMoreAboutDistributedGraphColoring


Tutorcircle.com
PageNo.:1/4

This is the result we had in the earlier section on differentiating trig funcions.
Derivative of 1/(sec x - tan x)?
lets simplify your equation..........
1/(sec x - tan x)
=(sec x + tan x) / (sec x - tan x)(sec x + tan x)
=(sec x + tan x) / (sec^2 - tan^2)
=(sec x + tan x) ............. (since sec^2 = 1 + tan^2 )
now derivatin above simplified equation........ sec x* tan x + sec^2
= sec x ( sec x + tan x )....... ans.
Proving the derivative of tan x?
Tan' (x)= (sec x)^2
lim (h0) [f(x + h) - f(x)]/h
= lim (h0) [tan(x + h) - tan(x)]/h
= lim (h0) [sin(x + h)/cos(x + h) - sin(x)/cos(x)]/h
= lim (h0) [(cos(x)sin(x + h) - sin(x)cos(x + h))] / [h cos(x)cos(x + h)]
= lim (h0) [cos(x)(sin(x)cos(h) + cos(x)sin(h)) - sin(x)(cos(x)cos(h) - sin(x)sin(h))] /
[h cos(x)cos(x + h)]
= lim (h0) [sin(x)cos(x)cos(h) + cos(x)sin(h) - sin(x)cos(x)cos(h) + sin(x)sin(h)) ] /
ReadMoreAboutPropertiesofsquareRoots


Tutorcircle.com
PageNo.:2/4

[h cos(x)cos(x + h)]
= lim (h0) [cos(x)sin(h) + sin(x)sin(h)) ] / [h cos(x)cos(x + h)]
= lim (h0) [sin(h)(cos(x) + sin(x))] / [h cos(x)cos(x + h)]
= lim (h0) sin(h) / [h cos(x)cos(x + h)]
= lim (h0) [sin(h)/h] * lim (h0) [1/cos(x)cos(x + h)]
= 1 * 1/ [cos(x)cos(x + 0)]
= 1/cos(x)
= sec(x)


Tut
Tu o
t rc
r i
c rc
r l
c e
l .
e c
. o
c m
Pa
P ge
g
e No
N ..::2/
3 3
/4

ThankYouForWatching
Presentation



Document Outline

  • ﾿