Differentiation of sec inverse x

Here you will learn differentiation of sec inverse x or arcsecx x by using chain rule.

Let’s begin –

Differentiation of sec inverse x or \(sec^{-1}x\) :

If x \(\in\) R – [-1, 1] . then the differentiation of \(sec^{-1}x\) with respect to x is \(1\over | x |\sqrt{x^2 – 1}\).

i.e. \(d\over dx\) \(sec^{-1}x\) = \(1\over | x |\sqrt{x^2 – 1}\).

Proof using chain rule :

Let y = \(sec^{-1}x\). Then,

\(sec(sec^{-1}x)\) = x

\(\implies\) sec y = x

Differentiating both sides with respect to x, we get

\(d\over dx\)(sec y) = \(d\over dx\)(x)

\(d\over dx\) (sec y) = 1

By chain rule,

sec y tan y \(dy\over dx\) = 1

\(dy\over dx\) = \(1\over sec y tan y\)

\(dy\over dx\) = \(1\over | sec y | | tan y |\)

\(dy\over dx\) = \(1\over | sec y | \sqrt{tan^2 y}\)

\(\implies\) \(dy\over dx\) = \(1\over | sec y | \sqrt{sec^2 y – 1}\)

\(\implies\) \(d\over dx\) \(sec^{-1}x\) = \(1\over | x |\sqrt{x^2 – 1}\)

Hence, the differentiation of \(sec^{-1}x\) with respect to x is \(1\over | x |\sqrt{x^2 – 1}\).

Example : What is the differentiation of \(sec^{-1} x^2\) with respect to x ?

Solution : Let y = \(sec^{-1} x^2\)

Differentiating both sides with respect to x and using chain rule, we get

\(dy\over dx\) = \(d\over dx\) (\(sec^{-1} x^2\))

\(dy\over dx\) = \(1\over | x^2 | \sqrt{x^4 – 1}\).(2x) = \(2x\over | x^2 | \sqrt{x^4 – 1}\)

Hence, \(d\over dx\) (\(sec^{-1} x^2\)) = \(2x\over | x^2 | \sqrt{x^4 – 1}\)

Example : What is the differentiation of x + \(sec^{-1} x\) with respect to x ?

Solution : Let y = x + \(sec^{-1} x\)

Differentiating both sides with respect to x, we get

\(dy\over dx\) = \(d\over dx\) (x) + \(d\over dx\) (\(sec^{-1} x\))

\(dy\over dx\) = 1 + \(1\over | x | \sqrt{x^2 – 1}\)

Hence, \(d\over dx\) (x + \(sec^{-1} x\)) = 1 + \(1\over | x | \sqrt{x^2 – 1}\)


Related Questions

What is the Differentiation of tan inverse x ?

What is the Differentiation of secx ?

What is the Integration of Sec Inverse x and Cosec Inverse x ?

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