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}$

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