Math Doubts

Derivative of tanx formula Proof

The derivative of tan function with respect to a variable is equal to square of secant. If $x$ is a variable, then the tangent function is written as $\tan{x}$. The differentiation of the $\tan{x}$ with respect to $x$ is equal to $\sec^2{x}$ and it can be proved mathematically by first principle.

Express Differentiation of function in Limit form

According to definition of the derivative, the derivative of the function in terms of $x$ can be written in the following limiting operation form.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

If $f{(x)} = \tan{x}$, then $f{(x+h)} = \tan{(x+h)}$. Now, the proof of the differentiation of $\tan{x}$ function with respect to $x$ can be started from first principle.

$\implies$ $\dfrac{d}{dx}{\, (\tan{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan{(x+h)}-\tan{x}}{h}}$

Simplify the entire function

The difference of the tan functions in the numerator can be simplified by the quotient rule of sine and cos functions.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{\sin{(x+h)}}{\cos{(x+h)}} -\dfrac{\sin{x}}{\cos{x}}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{\sin{(x+h)}\cos{x}-\cos{(x+h)}\sin{x}}{\cos{(x+h)}\cos{x}}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{(x+h)}\cos{x}-\cos{(x+h)}\sin{x}}{h\cos{(x+h)}\cos{x}}}$

The trigonometric expression in the numerator is the expansion of the angle difference identity of the sin function.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{(x+h-x)}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{(\cancel{x}+h-\cancel{x})}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{h}}{h\cos{(x+h)}\cos{x}}}$

Now, divide the limit of the function as the limit of product of two functions.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin{h}}{h}}$ $\times$ $\dfrac{1}{\cos{(x+h)}\cos{x}} \Bigg]$

Apply product rule of limits for evaluating limit of the product of the functions.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{h}}{h}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{\cos{(x+h)}\cos{x}}\Bigg)}$

Evaluate Limits of trigonometric functions

As per limit of sinx/x as x approaches 0 formula, the limit of the first trigonometric function is equal to one as $h$ tends to zero.

$=\,\,\,$ $1 \times \Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{\cos{(x+h)}\cos{x}}\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{\cos{(x+h)}\cos{x}}}$

Now, evaluate the limit of the trigonometric function by the direct substitution method.

$=\,\,\, \dfrac{1}{\cos{(x+0)}\cos{x}}$

$=\,\,\, \dfrac{1}{\cos{x}\cos{x}}$

$\implies$ $\dfrac{d}{dx}{\, (\tan{x})}$ $\,=\,$ $\dfrac{1}{\cos^2{x}}$

The reciprocal of the cosine function is equal to secant as per reciprocal identity of cos function.

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, (\tan{x})}$ $\,=\,$ $\sec^2{x}$

Thus, the derivative of tan function with respect to a variable is equal to square of secant function, is derived mathematically from first principle.

Math Doubts

A best free mathematics education website that helps students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

A math help place with list of solved problems with answers and worksheets on every concept for your practice.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved