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 ?
