Before going to find the derivative that sec x, let united state recall a few things. Sec x is the reciprocal of cos x and tan x is the ratio of sin x and also cos x. These meanings of sec x and tan x are really important to perform the differentiation that sec x with respect come x. We can find it using various ways such as:
by making use of the very first principleby using the quotient ruleby utilizing the chain ruleLet us perform the differentiation that sec x in every of this methods and also we will fix a couple of problems making use of the derivative that sec x.
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| 1. | What is Derivative that Sec x? |
| 2. | Derivative that Sec x by very first Principle |
| 3. | Derivative that Sec x through Quotient Rule |
| 4. | Derivative that Sec x by Chain Rule |
| 5. | FAQs top top Derivative of Sec x |
What is Derivative the Sec x?
The derivative of sec x v respect come x is sec x · tan x. I.e., that is the product of sec x and tan x. We represent the derivative of sec x with respect come x v d/dx(sec x) (or) (sec x)'. Thus,
d/dx (sec x) = sec x · tan x (or)(sec x)' = sec x · tan x
But wherein is tan x coming from in the derivative the sec x? We space going to differentiate sec x in various methods such as making use of the very first principles (definition of the derivative), quotient rule, and chain preeminence in the upcoming sections.
Derivative the Sec x by first Principle
We are going to prove that the derivative of sec x is sec x · tan x by making use of the first principles (or) the an interpretation of the derivative. For this, assume the f(x) = sec x.
Proof:
By first principle, the derivative the a function f(x) is,
f'(x) = limₕ→₀
Since f(x) = sec x, we have f(x + h) = sec (x + h).
Substituting these worths in (1),
f' (x) = limₕ→₀
= limₕ→₀ 1/h <1/(cos (x + h) - 1/cos x)>
= limₕ→₀ 1/h
By sum to product formulas, cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2. So
f'(x) = 1/cos x limₕ→₀ 1/h <- 2 sin (x + x + h)/2 sin (x - x - h)/2> /
= 1/cos x limₕ→₀ 1/h <- 2 sin (2x + h)/2 sin (- h)/2> /
Multiply and divide through h/2,
= 1/cos x limₕ→₀ (1/h) (h/2) <- 2 sin (2x + h)/2 sin (- h/2) / (h/2)> /
When h → 0, we have 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)
We have limₓ→₀ (sin x) / x = 1. So
f'(x) = 1/cos x. 1. Sin x/cos x
We recognize that 1/cos x = sec x and also sin x/cos x = tan x. So
f'(x) = sec x · tan x
Hence proved.
Derivative that Sec x by Quotient Rule
We will prove that the differentiation of sec x with respect come x offers sec x · tan x by making use of the quotient rule. Because that this, we will certainly assume the f(x) = sec x and it deserve to be created as f(x) = 1/cos x.
Proof:
We have actually f(x) = 1/cos x = u/v
By quotient rule,
f'(x) = (vu' - uv') / v2
f'(x) =
=
= (sin x) / cos2x
= 1/cos x · (sin x)/(cos x)
= sec x · tan x
Hence proved.
Derivative that Sec x through Chain Rule
To prove that the derivative that sec x to be sec x · tan x by chain rule, we will assume that f(x) = sec x = 1/cos x.
Proof:
We deserve to write f(x) as,
f(x) = 1/cos x = (cos x)-1
By power rule and also chain rule,
f'(x) = (-1) (cos x)-2 d/dx(cos x)
By a residential property of exponents, a-m = 1/am. Also, we recognize that d/dx(cos x) = - sin x. So
f'(x) = -1/cos2x · (- sin x)
= (sin x) / cos2x
= 1/cos x · (sin x)/(cos x)
= sec x · tan x
Hence proved.
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