Derivative of secx

In the previous post we have discussed about cosine law and In today's session we are going to discuss about Derivative of secx. Trigonometric can be defined as the relationship between the angles and sides of a triangle. In the trigonometric there are different types of function and derivative of functions. Here we will study the different types of derivatives. Derivatives of trigonometric  function are mention below:

d / da sin (a) = cos (a)	d / da cos (a) = - sin (a)
d / da tan (a) = sec2 (a)	d / da csc (a) = - csc (a) cot (a)
d / da sec (a) = sec (a) tan (a)	d / da cot (a) = - csc2 (a)

(know more about Derivative, here). These above mention are the different types of derivatives of trigonometric functions. Now we will find the Derivative of secx. We are earlier studies the derivative of sec x is sec (x) tan (x). Now see the prove of Derivative Of Secx. To find the derivative of sec x first we write the sec x in the derivative form:

Prove = d / dx sec = sec x tan x;

We know that sec x = 1 / cos x;

We can also write in place of sec x as:

= d / dx [1 / cos x]

We can solve it by u / v methods:

u / v = u d / dx (v) – v d / dx u / u2

So we can write the above expression as:

= [cos x d / dx (1) – 1 d / dx cos x] / cos2 x;

If we find the derivative of 1 and cos x we get:

= [cos x (0) – 1 (- sin x)] / cos2x;

If we solve we get:

= [0 + sin x] / cos2 x;

On further solving we get:

= Sec x tan x.

So the proof of sec x is sec x tan x.

This is how we can solve the derivative value. Centripetal Acceleration Formula is used to find the speed along to a given circular path and its radius is directed along to the center. icse sample papers 2013 is very helpful for exam point of view.