Proof of Derivative of Sec x
The derivative of sec x can be proved using the following ways:
- By using the First Principle of Derivative
- By using Quotient Rule
- By using Chain Rule
Derivative of Sec x by First Principle of Derivative
To prove derivative of sec x using First Principle of Derivative, we will use basic limits and trigonometric formulas which are listed below:
- cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2.
- limx→0 (sin x) / x = 1
- 1/cos x = sec x
- sin x/cos x = tan x.
Let’s start the proof for the derivative of sec x ,assume that f(x) = sec x.
By first principle, the derivative of a function f(x) is,
f'(x) = limh→0[f(x + h) – f(x)] / h … (1)
Since f(x) = sec x, we have f(x + h) = sec (x + h).
Substituting these values in (1),
f’ (x) = limh→0 [sec (x + h) – sec x]/h
⇒ limh→0 1/h [1/(cos (x + h) – 1/cos x)]
⇒limh→0 1/h [cos x – cos(x + h)] / [cos x cos(x + h)]
⇒ 1/cos x limh->0 1/h [- 2 sin (x + x + h)/2 sin (x – x – h)/2] / [cos(x + h)] {By 1}
⇒ 1/cos x limh->0 1/h [- 2 sin (2x + h)/2 sin (- h)/2] / [cos(x + h)]
Multiply and divide by h/2,
⇒ 1/cos x limh->0 (1/h) (h/2) [- 2 sin (2x + h)/2 sin (- h/2) / (h/2)] / [cos(x + h)]
When h → 0, we have h/2 → 0. So,
⇒ 1/cos x Lim h/2->0 sin (h/2) / (h/2). limh->0(sin(2x + h)/2)/cos(x + h)
⇒ 1/cos x. 1. sin x/cos x {By 2}
⇒ sec x · tan x {By 3 & 4}
Therefore, f'(x) = d/dx [sec x] = sec x . tan x
Derivative of Sec x by Quotient Rule
To prove derivative of sec x using Quotient rule, we will use basic derivatives and trigonometric formulas which are listed below:
- sec x = 1/cos x
- (d/dx) [u/v] = [u’v – uv’]/v2
Let’s start the proof of the derivative of sec x, assume that f(x) = sec x = 1/cos x.
We have f(x) = 1/cos x = u/v
By quotient rule,
f'(x) = (vu’ – uv’) / v2
f'(x) = [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)
⇒ sec x · tan x
Therefore, f'(x) = d/dx [sec x] = sec x. tan x
Derivative of Sec x by Chain Rule
To prove derivative of sin x using chain rule, we will use basic derivatives and trigonometric formulas which are listed below:
- a-m = 1/am
- d/dx [cos x] = – sin x
- d/dx [xn] = nxn-1
Let’s start the proof of the derivative of sec x, assume that f(x) = sec x = 1/cos x.
We can write f(x) as,
f(x) = 1/cos x = (cos x)-1
By power rule and chain rule,
f'(x) = (-1) (cos x)-2 d/dx (cos x) {By 3}
⇒ -1/cos2x · (- sin x) {By 1 & 2}
⇒ (sin x) / cos2x
⇒ 1/cos x · (sin x)/ (cos x)
⇒ sec x · tan x
Therefore, f'(x) = d/dx [sec x] = sec x. tan x
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Derivative of Sec x
Derivative of Sec x is sec x tan x. Derivative of Sec x refers to the process of finding the change in the secant function with respect to the independent variable. The specific process of finding the derivative for trigonometric functions is referred to as trigonometric differentiation, and the derivative of Sec x is one of the key results in trigonometric differentiation.
In this article, we will learn about the derivative of sec x and its formula including the proof of the formula using the first principle of derivatives, quotient rule, and chain rule as well.