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.
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.
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