Here you will learn what is the differentiation of cosecx and its proof by using first principle.
Let’s begin –
Differentiation of cosecx
The differentiation of cosecx with respect to x is -cosecx.cotx
i.e. \(d\over dx\) (cosecx) = -cosecx.cotx
Proof Using First Principle :
Let f(x) = cosec x. Then, f(x + h) = cosec(x + h)
\(\therefore\) \(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(f(x + h) – f(x)\over h\)
\(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(cosec(x + h) – cosec x\over h\)
\(\implies\) \(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \({1\over sin(x + h)} – {1\over sin x}\over h\)
\(\implies\) \(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(sin x – sin(x + h)\over h sin x sin(x +h)\)
By using trigonometry formula,
[sin C – sin D = \(2sin ({C – D\over 2})cos ({C + D\over 2})\)]
\(\implies\) \(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(2sin ({x – x – h\over 2})cos({x + x + h\over 2})\over h sin x sin (x + h)\)
\(\implies\) \(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(2sin ({-h\over 2})cos({x + h/2})\over h sin x sin (x + h)\)
\(\implies\) \(d\over dx\)(f(x)) = -\(lim_{h\to 0}\) \(cos ({x + h/2})\over sin x sin(x + h)\).\(lim_{h\to 0}\) \(sin(h/2)\over (h/2)\)
because, [\(lim_{h\to 0}\)\(sin(h/2)\over (h/2)\) = 1]
\(\implies\) \(d\over dx\)(f(x)) = -\(cos x\over sin x sin x\)(1) = -cot x cosec x
Hence, \(d\over dx\) (cosec x) = =cosecx.cotx
Example : What is the differentiation of cosec x + x with respect to x?
Solution : Let y = cosec x + x
\(d\over dx\)(y) = \(d\over dx\)(cosec x + x)
\(\implies\) \(d\over dx\)(y) = \(d\over dx\)(cosec x) + \(d\over dx\)(x)
By using cosecx differentiation we get,
\(\implies\) \(d\over dx\)(y) = -cosec x cot x + 1
Hence, \(d\over dx\)(sec x + x) = -cosec x cot x + 1
Example : What is the differentiation of \(cosec\sqrt{x}\) with respect to x?
Solution : Let y = \(cosec\sqrt{x}\)
\(d\over dx\)(y) = \(d\over dx\)(\(cosec\sqrt{x}\))
By using chain rule we get,
\(\implies\) \(d\over dx\)(y) = \(1\over 2\sqrt{x}\)(\(-cosec \sqrt{x}.cot\sqrt{x}\))
Hence, \(d\over dx\)(\(cosec\sqrt{x}\)) = -\(1\over 2\sqrt{x}\)(\(cosec \sqrt{x}.cot\sqrt{x}\))
Related Questions
What is the Differentiation of cosec inverse x ?