Here you will learn differentiation of tan inverse x or arctanx x by using chain rule.
Let’s begin –
Differentiation of tan inverse x or \(tan^{-1}x\) :
The differentiation of \(tan^{-1}x\) with respect to x is \(1\over {1 + x^2}\).
i.e. \(d\over dx\) \(tan^{-1}x\) = \(1\over {1 + x^2}\).
Proof using chain rule :
Let y = \(tan^{-1}x\). Then,
\(tan(tan^{-1}x)\) = x
\(\implies\) tan y = x
Differentiating both sides with respect to x, we get
\(d\over dx\)(tan y) = \(d\over dx\)(x)
\(d\over dx\) (tan y) = 1
By chain rule,
\(sec^2 y\) \(dy\over dx\) = 1
\(dy\over dx\) = \(1\over sec^2 y\)
[ \(\because\) 1 + \(tan^2 y\) = \(sec^2 y\)
\(dy\over dx\) = \(1\over {1 + tan^2 y}\)
\(\implies\) \(dy\over dx\) = \(1\over {1 + x^2}\)
\(\implies\) \(d\over dx\) \(tan^{-1}x\) = \(1\over {1 + x^2}\)
Hence, the differentiation of \(tan^{-1}x\) with respect to x is \(1\over {1 + x^2}\).
Example : What is the differentiation of \(tan^{-1} x^2\) with respect to x ?
Solution : Let y = \(tan^{-1} x^2\)
Differentiating both sides with respect to x and using chain rule, we get
\(dy\over dx\) = \(d\over dx\) (\(tan^{-1} x^2\))
\(dy\over dx\) = \(1\over {1 + x^4}\).(2x) = \(2x\over {1 + x^4}\)
Hence, \(d\over dx\) (\(tan^{-1} x^2\)) = \(2x\over {1 + x^4}\)
Example : What is the differentiation of 2x + \(tan^{-1} x\) with respect to x ?
Solution : Let y = 2x + \(tan^{-1} x\)
Differentiating both sides with respect to x, we get
\(dy\over dx\) = \(d\over dx\) (2x) + \(d\over dx\) (\(tan^{-1} x\))
\(dy\over dx\) = 2 + \(1\over {1 + x^2}\)
Hence, \(d\over dx\) (2x + \(tan^{-1} x\)) = 2 + \(1\over {1 + x^2}\)
Related Questions
What is the Differentiation of tanx ?