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.

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.

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

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.

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

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.

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