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