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