Here you will learn differentiation of sin inverse x or arcsin x by using chain rule.
Let’s begin –
Differentiation of sin inverse x or \(sin^{-1}x\) :
If x \(\in\) (-1, 1) , then the differentiation of \(sin^{-1}x\) with respect to x is \(1\over \sqrt{1 – x^2}\).
i.e. \(d\over dx\) \(sin^{-1}x\) = \(1\over \sqrt{1 – x^2}\) , for x \(\in\) (-1, 1).
Proof using chain rule :
Let y = \(sin^{-1}x\). Then,
\(sin(sin^{-1}x)\) = x \(\implies\) sin y = x
Differentiating both sides with respect to x, we get
1 = \(d\over dx\) (sin y)
By chain rule,
1 = \(d\over dx\) (sin y) \(\times\) \(dy\over dx\)
1 = cos y \(dy\over dx\)
\(dy\over dx\) = \(1\over cos y\)
\(dy\over dx\) = \(1\over \sqrt{1 – sin^2 y}\)
\(\implies\) \(dy\over dx\) = \(1\over \sqrt{1 – x^2}\)
\(\implies\) \(d\over dx\) \(sin^{-1}x\) = \(1\over \sqrt{1 – x^2}\)
Hence, the differentiation of \(sin^{-1}x\) with respect to x is \(1\over \sqrt{1 – x^2}\).
Example : What is the differentiation of \(sin^{-1} x^3\) with respect to x ?
Solution : Let y = \(sin^{-1} x^3\)
Differentiating both sides with respect to x and using chain rule, we get
\(dy\over dx\) = \(d\over dx\) (\(sin^{-1} x^3\))
\(dy\over dx\) = \(1\over \sqrt{1 – x^6}\).\(3x^2\) = \(3x^2\over \sqrt{1 – x^6}\)
Hence, \(d\over dx\) (\(sin^{-1} x^3\)) = \(3x^2\over \sqrt{1 – x^6}\)
Example : What is the differentiation of \(x^2\) + \(sin^{-1} x^5\) with respect to x ?
Solution : Let y = \(x^2\) + \(sin^{-1} x^5\)
Differentiating both sides with respect to x and using chain rule, we get
\(dy\over dx\) = \(d\over dx\) (\(x^2\)) + \(d\over dx\) (\(sin^{-1} x^5\))
\(dy\over dx\) = 2x + \(1\over \sqrt{1 – x^{10}}\).\(5x^4\)
Hence, \(d\over dx\) (\(x^2\) + \(sin^{-1} x^5\))= 2x + \(5x^4\over \sqrt{1 – x^{10}}\)
Related Questions
What is the Differentiation of sin x ?