In this article, you will learn what is the derivative of sec x as well as prove the derivative of sec x by quotient rule, first principal rule, and chain rule.
I have already discussed the derivative of sin x, cos x, and tan x in a previous article you can check from here.
So, without wasting time let's get started.
What is sec x?
Sec x is a trigonometric function that is reciprocal of cos x.
Derivative of sec x
The derivative of sec x is equal to the product of sec x and tan x.
We can prove the derivative of sec x in three ways first by using the quotient rule and second by using the first principle rule and the last chain rule.
Derivative of sec x Proof by Quotient Rule
The formula of the quotient rule is,
dy/dx = {v (du/dx) - u (dv/dx)}/v²
Where,
dy/dx = derivative of y with respect to x
v = variable v
du/dx = derivative of u with respect to x
u = variable u
dv/dx = derivative of v with respect to x
v = variable v
Let us,
y = sec x
As we know,
sec x = 1/cos x
So,
We can written as,
y = 1/cos x.
Where,
u = 1
v = cos x
Now putting these values on the quotient rule formula, we will get
dy/dx = [cos x d/dx(1) - 1 d/dx(cos x)] / (cos x)²
= [cos x (0) - 1 (-sin x)] / (cos²x)
= (sin x) / (cos²x)
= (1/cos x) × {(sin x)/(cos x)}
So, as we know,
1/cos x = sec x and sin x/cos x = tan x
Hence, after putting these values we will get,
d/dx (sec x) = sec x · tan x
Thus, we proved the derivative of sec x will be equal to sec x · tan x using the quotient rule method.
Derivative of sec x Proof by First Principle Rule
According to the first principle rule, the derivative limit of a function can be determined by computing the formula:
For a differentiable function y = f (x)
We define its derivative w.r.t x as :
dy/dx = f ' (x) = limₕ→₀ [f(x+h) - f(x)]/h
f'(x) = limₕ→₀ [f(x+h) - f(x)]/h
This limit is used to represent the instantaneous rate of change of the function f(x).
Let,
f (x) = sec x
So,
f(x + h) = sec (x + h)
Putting these values on the above first principle rules equation.
f' (x) = limₕ→₀ [sec (x + h) - sec x]/h
So, as we know sec x = 1/cos x
Hence, after putting 1/cos x in place of sec x we will get,
= limₕ→₀ 1/h [1/(cos (x + h) - 1/cos x)]
After solving this we will get,
= limₕ→₀ 1/h [cos x - cos(x + h)] / [cos x cos(x + h)]
So, as we know
cos a - cos b = -2 [{sin (a + b)/ 2} × {sin (a - b)/2}]
Now putting these values,
f'(x) = 1/cos x limₕ→₀ 1/h [- 2 sin (x + x + h)/2 × sin (x - x - h)/2] / [cos(x + h)]
= 1/cos x limₕ→₀ 1/h [- 2 sin (2x + h)/2 sin (- h)/2] / [cos(x + h)]
Now multiply and divide by h/2 we will get,
= 1/cos x limₕ→₀ (1/h) (h/2) [- 2 sin (2x + h)/2 sin (- h/2) / (h/2)] / [cos(x + h)]
If h→₀ then also h/2→₀
So
f'(x) = 1/cos x limₕ/₂→₀ sin (h/2) / (h/2) × limₕ→₀ (sin(2x + h)/2)/cos(x + h)
So, for limₓ→₀ sin x / x = 1
Hence,
f'(x) = 1/cos x × 1 × sin x/cos x
As we know,
1/cos x = sec x
sin x/cos x = tan x
After putting these values we will get,
f'(x) = sec x · tan x
Thus, we proved the derivative of sec x will be equal to sec x · tan x using the first principle rule method.
Derivative of sec x Proof by Chain Rule
Let us,
y = sec x
As we know,
sec x = 1/cos x
So,
y = 1 / (cos x)
= (cos x)⁻¹
By using the chain rule,
The formula of chain rule is,
dy/dx = (dy/du) × (du/dx)
Where,
dy/dx = derivative of y with respect to x
dy/du = derivative of y with respect to u
du/dx = derivative of u with respect to x
After putting these values we can find,
dy/dx = (-1) (cos x)⁻² d/dx(cos x)
Since,
d/dx(cos x) = - sin x and a⁻ⁿ = 1/aⁿ
Hence,
dy/dx = -1/cos²x · (- sin x)
= (sin x) / cos²x
= 1/cos x · (sin x)/(cos x)
d/dx (sec x) = sec x · tan x
Thus, we proved the derivative of sec x will be equal to sec x · tan x using the chain rule method.
See Also:
FAQ Related to the Derivative of sec x
What is the derivative of sec(x)?
The derivative of sec x is equal to the product of sec x and tan x.
Is secant the inverse of cosine?
Yes, secant is the inverse of cosine.
What is the derivative of the sec inverse of x?
The derivative of the sec inverse of x is the 1/(x √x² - 1).
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