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.

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

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

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

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
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more