Here you will learn differentiation of cos inverse x or arccos x by using chain rule.
Let’s begin –
Differentiation of cos inverse x or \(cos^{-1}x\) :
If x \(\in\) (-1, 1) , then the differentiation of \(cos^{-1}x\) with respect to x is \(-1\over \sqrt{1 – x^2}\).
i.e. \(d\over dx\) \(cos^{-1}x\) = \(-1\over \sqrt{1 – x^2}\) , for x \(\in\) (-1, 1).
Proof using chain rule :
Let y = \(cos^{-1}x\). Then,
\(cos(cos^{-1}x)\) = x
\(\implies\) cos y = x
Differentiating both sides with respect to x, we get
\(d\over dx\)(cos y) = \(d\over dx\)(x)
\(d\over dx\) (cos y) = 1
By chain rule,
-sin y \(dy\over dx\) = 1
\(dy\over dx\) = \(-1\over sin y\)
\(dy\over dx\) = \(-1\over \sqrt{1 – cos^2 y}\)
\(\implies\) \(dy\over dx\) = \(-1\over \sqrt{1 – x^2}\)
\(\implies\) \(d\over dx\) \(cos^{-1}x\) = \(-1\over \sqrt{1 – x^2}\)
Hence, the differentiation of \(cos^{-1}x\) with respect to x is \(-1\over \sqrt{1 – x^2}\).
Example : What is the differentiation of \(cos^{-1} x^3\) with respect to x ?
Solution : Let y = \(cos^{-1} x^3\)
Differentiating both sides with respect to x and using chain rule, we get
\(dy\over dx\) = \(d\over dx\) (\(cos^{-1} x^3\))
\(dy\over dx\) = \(-1\over \sqrt{1 – x^6}\).\(3x^2\) = \(-3x^2\over \sqrt{1 – x^6}\)
Hence, \(d\over dx\) (\(cos^{-1} x^3\)) = \(-3x^2\over \sqrt{1 – x^6}\)
Example : What is the differentiation of x + \(cos^{-1} x\) with respect to x ?
Solution : Let y = x + \(cos^{-1} x\)
Differentiating both sides with respect to x, we get
\(dy\over dx\) = \(d\over dx\) (x) + \(d\over dx\) (\(cos^{-1} x\))
\(dy\over dx\) = 1 + \(-1\over \sqrt{1 – x^2}\)
Hence, \(d\over dx\) (x + \(cos^{-1} x\)) = 1 – \(1\over \sqrt{1 – x^2}\)
Related Questions
What is the Differentiation of cos x ?