The differentiation of tan function can be derived in calculus by the limit rules and trigonometric identities. The following step by step procedure helps you to learn how to derive derivative of tan function mathematically.

The derivative of a function can be derived in limit form by the principle relation of differentiation of a function in limit form.

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

Take $f(x) = \tan x$ then $f(x+h) = \tan(x+h)$. Substitute these two tan functions and let us start deriving the differentiation formula of trigonometric function tangent.

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

Each tangent function can be expressed in terms of sine and cosine functions as per quotient relation of sine and cosine functions with tan function. It helps us to simplify this mathematical expression easily.

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

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

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

The trigonometric expression in numerator represents sine of compound angle identity. Therefore, it can be minimized by applying sine compound angle formula.

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

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

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

The expression can be simplified mathematically by splitting it as two multiplying factors.

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

Apply the limit rule for the product of the both functions.

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

According to lim sinx/x as x approaches 0 formula, the value of the first limit function is one and substitute $h = 0$ in the second multiplying limit function.

$\implies \dfrac{d}{dx} \, \tan{x}$ $\,=\,$ $1\times \dfrac{1}{\cos{(x+0)}\cos{x}}$

$\implies \dfrac{d}{dx} \, \tan{x}$ $\,=\,$ $\dfrac{1}{\cos{x}\cos{x}}$

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

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

It is proved that the derivative of $\tan{x}$ with respect to $x$ is equal to square of $\sec{x}$.

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