Derivative of Arcsec

The derivative of arcsec is equal to 1/[|x| √(x² - 1)]. Arcsec is the inverse of the secant function and is one of the important inverse trigonometric functions denoted by sec^-1x (read as sec inverse x) or arcsec x. Therefore, the formula for the derivative of sec inverse x can be written as d(sec^-1x)/dx = 1/[|x| √(x² - 1)] = d(arcsec x)/dx. Now, the value of x lies in the union of the intervals (-∞, -1) and (1, ∞) as the derivative of arcsec is not defined in the interval [-1, 1].

Further in this article, we will explore the derivative of sec inverse x and derive its formula. We will also understand the reason behind the absolute sign in the derivative of arcsec with the help of the graph of arcsec along with a few solved examples for a better understanding of the concept.

The derivative of arcsec gives the slope of the tangent to the graph of the inverse secant function. The formula for the derivative of sec inverse x is given by d(arcsec)/dx = 1/[|x| √(x² - 1)]. This derivative is also denoted by d(sec^-1x)/dx. We can evaluate the derivative of arcsec by assuming arcsec to be equal to some variable and then taking the inverse sec on both sides. The value of x in the derivative of arcsec lies in the interval (-∞, -1) U (1, ∞) because the values of the interval [-1, 1] make the derivative undefined. The next section of the article discusses the formula of the derivative of sec inverse x.

The formula for the derivative of sec inverse x is given by d(sec^-1x)/dx = 1/[|x| √(x² - 1)], where x belongs to the intervals (-∞, -1) and (1, ∞). If the value of x = 1 or -1, then the denominator of the derivative becomes zero and for values of x in the interval (-1, 1), the value inside the square root in the denominator of the derivative of arcsec becomes negative. Hence, the value of x does not belong to the interval [-1, 1] for the derivative of sec inverse x.

Now that we know that the derivative of arcsec is equal to d(sec^-1x)/dx = 1/[|x| √(x² - 1)], we will prove it using the trigonometric formulas. Assume y = sec^-1x, then using the definition of sec inverse x, we can write it as sec y = x. Now differentiating sec y = x with respect to x, we have

sec y tan y × dy/dx = 1 --- [Because the derivative of secx is equal to secx tanx]

⇒ dy/dx = 1/(tan y sec y) --- (1)

Using the trigonometric identity 1 + tan²A = sec²A, we have tan²y = sec²y - 1 which implies tany = √(sec²y - 1) = √(x² - 1) [Because sec y = x]. Substituting these values in (1), we have

dy/dx = 1 / [x √(x² - 1)]

As we can see there is no absolute sign in the formula dy/dx = 1 / [x √(x² - 1)]. Now, we will understand the reason behind the absolute sign in the derivative of arcsec. The image given below shows the graph of the arcsec function. As we can see in the graph, the tangents of the graph have positive slopes for all values of x in the intervals (-∞, -1) and (1, ∞). But the sign of the derivative dy/dx = 1 / [x √(x² - 1)] depends on the value of x in the denominator as the square root can never be negative. Therefore, we have considered the absolute value of x in the denominator to get a positive slope (given by the derivative). Hence, the derivative of arcsec is equal to 1 / [|x| √(x² - 1)].

To derive the differentiation of sec inverse x using the first principle of derivatives, we will use certain inverse trigonometric, trigonometric, and limits formulas and identities. Now, assume y = sec^-1x which implies sec y = x. Using the trigonometric identity 1 + tan²A = sec²A, we can write tan²y = sec²y - 1 which implies tan y = √(sec²y - 1) = √(x² - 1) ⇒ y = tan^-1[√(x² - 1)] ⇒ sec^-1x = tan^-1[√(x² - 1)] [Because sec y = x]. We will also use the formula:

• sec^-1x = sec^-1x , if x ≥ 1 --- (1)
• sec^-1x = π - sec^-1x , if x ≤ -1 --- (2)
• f'(x) = lim_h→0[f(x+h) - f(x)]/h
• lim_x→0 (tan x)/x = 1
• tan^-1a - tan^-1b = tan^-1 [(a - b) / (1 + ab)]

First, we will consider x ≥ 1. Using the above formulas, we have

d(arcsec)/dx = lim_h→0[sec^-1(x+h) - sec^-1(x)]/h

= lim_h→0[tan^-1[√((x+h)² - 1)] - tan^-1[√(x² - 1)]]/h --- [Because we have considered x ≥ 1]

= lim_h→0 tan^-1{ [√((x+h)² - 1) - √(x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] }/h --- [Using tan^-1a - tan^-1b formula]

= lim_h→0 (1/h) × lim_h→0 tan^-1{ [√((x+h)² - 1) - √(x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] } / { [√((x+h)² - 1) - √(x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] } × lim_h→0 [√((x+h)² - 1) - √(x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] } --- {Mutiplying and dividing the limit by [√((x+h)² - 1) - √(x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] }

= lim_h→0 (1/h) × 1 × lim_h→0 [√((x+h)² - 1) - √(x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] } --- [Using lim_x→0 (tan x)/x = 1]

= lim_h→0 (1/h) × lim_h→0 [√((x+h)² - 1) - √(x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] }

= lim_h→0 (1/h) × lim_h→0 [√((x+h)² - 1) - √(x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] } × {[√((x+h)² - 1) + √(x² - 1)]/[√((x+h)² - 1) + √(x² - 1)]} --- [Multiplying and dividing by [√((x+h)² - 1) + √(x² - 1)]]

= lim_h→0 (1/h) × lim_h→0 [(x+h)² - 1 - (x² - 1)] / [1 + √((x+h)² - 1) √(x² - 1)] } × 1 / [√((x+h)² - 1) + √(x² - 1)]

= lim_h→0 (1/h) × lim_h→0 [(x+h)² - x²] / [1 + √((x+h)² - 1) √(x² - 1)] } × 1 / [√((x+h)² - 1) + √(x² - 1)]

= lim_h→0 (1/h) × lim_h→0 [(x+h+x) (x+h-x)] / [1 + √((x+h)² - 1) √(x² - 1)] } × 1 / [√((x+h)² - 1) + √(x² - 1)]

= lim_h→0 (1/h) × lim_h→0 [(x+h+x) h] / [1 + √((x+h)² - 1) √(x² - 1)] } × 1 / [√((x+h)² - 1) + √(x² - 1)]

= lim_h→0 (h/h) × lim_h→0 [(x+h+x)] / [1 + √((x+h)² - 1) √(x² - 1)] } × 1 / [√((x+h)² - 1) + √(x² - 1)]

= lim_h→0 [(2x+h)] / [1 + √((x+h)² - 1) √(x² - 1)] } × 1 / [√((x+h)² - 1) + √(x² - 1)]

= 2x / [1 + √(x² - 1) √(x² - 1)] } × 1 / [√(x² - 1) + √(x² - 1)]

= 2x / (1 + x² - 1) × 1 / 2 √(x² - 1)

= 2x / x² 2 √(x² - 1)

= 1 / (x √(x² - 1)) --- (A)

Now, 1 / (x √(x² - 1)) is the derivative of arcsec for x ≥ 1. Next, for x ≤ -1, we have

d(arcsec)/dx = lim_h→0[sec^-1(x+h) - sec^-1(x)]/h

= lim_h→0[π - tan^-1[√((x+h)² - 1)] - (π - tan^-1[√(x² - 1)])]/h --- [Because we have considered x ≤ -1]

= lim_h→0[π - tan^-1[√((x+h)² - 1)] - π + tan^-1[√(x² - 1)]]/h

= lim_h→0[ - tan^-1[√((x+h)² - 1)] + tan^-1[√(x² - 1)]]/h

= - lim_h→0[tan^-1[√((x+h)² - 1)] - tan^-1[√(x² - 1)]]/h

= - 1 / (x √(x² - 1)) --- [Using the derivation of (A)]

= 1 / (-x √(x² - 1)) --- (B)

Hence, 1 / (-x √(x² - 1)) is the derivative of sec inverse x for x ≤ -1. Therefore, combining (A) and (B), the derivative of arcsec is equal to 1 / [|x| √(x² - 1)], for all values of x in (-∞, -1) U (1, ∞). Hence, the differentiation of arcsec has been derived using the first principle of differentiation.

Important Notes on Derivative of Arcsec

• The derivative of arcsec is equal to 1 / [|x| √(x² - 1)].
• The absolute sign in the derivative of sec inverse x is because the tangents to the sec inverse graph have a positive slope.
• The differentiation of sec inverse is defined for values in (-∞, -1) U (1, ∞).

☛ Related Topics:

• Cot Inverse x
• Derivative of sin inverse
• Inverse Cosine

What is Derivative of Arsec in Calculus?

The derivative of arcsec gives the slope of the tangent to the graph of the inverse secant function. The formula for the derivative of sec inverse x is given by d(arcsec)/dx = 1/[|x| √(x² - 1)]. The value of x lies in the intervals (-∞, -1) and (1, ∞).

How to Find the Derivative of Arcsec?

We can find the derivative of arcsec using the substitution method. Assume y = sec^-1x, then using the definition of sec inverse x, we can write it as sec y = x. Now differentiating sec y = x with respect to x, we can find the value of dy/dx. dy/dx will give the value of the derivative of sec inverse x.

What is the Formula of Derivative of Sec Inverse?

The formula for the derivative of sec inverse x is given by d(sec^-1x)/dx = 1/[|x| √(x² - 1)], where x belongs to the intervals (-∞, -1) and (1, ∞).

Why is There an Absolute Value in the Derivative of Arcsec?

There is no absolute sign in the formula dy/dx = 1 / [x √(x² - 1)]. In the graph of arcsec, the tangents of the graph have positive slopes for all values of x in the intervals (-∞, -1) and (1, ∞). But the sign of the derivative dy/dx = 1 / [x √(x² - 1)] depends on the value of x in the denominator as the square root can never be negative. Therefore, we take the absolute value of x in the denominator to get a positive slope (given by the derivative). Hence, the derivative of arcsec is equal to 1 / [|x| √(x² - 1)].

Find the Derivative of Arcsec 4x.

The derivative of arcsec can be calculated using the chain rule. Therefore, we have d(arcsec 4x)/dx = d(arcsec 4x)/d(4x) × d(4x)/dx = 1 / [|4x| √((4x)² - 1)] × 4 = 4 /[|4x| √(16x² - 1)] = 1 / |x| √(16x² - 1). Therefore, the derivative of arcsec 4x is 1 / |x| √(16x² - 1).

What is the Domain of the Derivative of Sec Inverse Function?

The value of x in the derivative of arcsec d(sec^-1x)/dx = 1/[|x| √(x² - 1)] lies in the union of the interval (-∞, -1) U (1, ∞).

Why Does the Value of x Not Belong to [-1, 1] in Derivative of Sec Inverse?

The derivative of sec inverse x is d(sec^-1x)/dx = 1/[|x| √(x² - 1)]. If the value of x = 1 or -1, then the denominator of the derivative becomes zero and for values of x in the interval (-1, 1), the value inside the square root in the denominator of the derivative of arcsec becomes negative. Hence, the value of x does not belong to the interval [-1, 1] for the derivative of sec inverse x.