Differentiation of tanx

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 ?

What is the Differentiation of sec x ?

What is the Integration of Tan x ?

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