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