# Derivative of secx Proof

The derivative of secant function with respect to a variable is equal to the product of secant and tangent functions. If $x$ is considered to represents a variable, then the secant function is written in mathematical form as $\sec{x}$. The differentiation of the $\sec{x}$ with respect to $x$ is equal to the product of $\sec{x}$ and $\tan{x}$. According to the first principle of differentiation, the derivative of secant function is derived in differential calculus.

### Differentiation of function in Limit form

Firstly, write the derivative of a function in terms of $x$ in limits operation form according to the definition of the derivative.

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

Now, take $\Delta x = h$ and write the equation in terms of $h$ instead of $\Delta x$.

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

If $f{(x)} = \sec{x}$, then $f{(x+h)} = \sec{(x+h)}$. Now, let’s start proving the differentiation of $\sec{x}$ function with respect to $x$ by first principle.

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

### Simplify the entire function

According to reciprocal identity of cosine function, the difference of the secant functions in the numerator can be simplified by converting them into cosine functions .

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

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

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

The difference of cosine functions can be combined by using difference to product trigonometric identity of cos functions.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{x+x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{x-(x+h)}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{x-x-h}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\require{cancel} \large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{\cancel{x}-\cancel{x}-h}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{-h}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

The sine of negative angle is equal to negative of sine of angle as per even/ odd trigonometric identity of sine function.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\Bigg(-\sin{\Bigg[\dfrac{h}{2}\Bigg]\Bigg)}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

It is essential to split the trigonometric function as product of two functions.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Bigg[\dfrac{2x+h}{2}\Bigg]} \times \sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h \times \cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]} \times 2\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\cos{(x+h)}\cos{x} \times h}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\cos{(x+h)}\cos{x}} \times \dfrac{2\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h}\Bigg)}$

By using product rule of limits, the limit of product of two functions can be evaluated by calculating the product of their limits.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\cos{(x+h)}\cos{x}}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h}}$

In numerator of second factor, there is a number $2$. It actually multiplies the other factor and it divides the function in denominator. So, shift it to denominator.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\cos{(x+h)}\cos{x}}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

### Evaluate Limits of trigonometric functions

Let us first find the limit of the first trigonometric function by the direct substitution method but do not disturb the second function.

$=\,\,\,$ $\dfrac{\sin{\Bigg[\dfrac{2x+0}{2}\Bigg]}}{\cos{(x+0)}\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\dfrac{\sin{\Bigg[\dfrac{2x}{2}\Bigg]}}{\cos{x}\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\require{cancel} \dfrac{\sin{\Bigg[\dfrac{\cancel{2}x}{\cancel{2}}\Bigg]}}{\cos{x}\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\dfrac{\sin{x}}{\cos{x}\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\dfrac{\sin{x}}{\cos{x}} \times \dfrac{1}{\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\tan{x} \times \sec{x}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\tan{x}\sec{x}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\sec{x}\tan{x}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

The second function is similar to the limit of $\dfrac{\sin{x}}{x}$ as $x$ approaches $0$ formula but the input is slightly different. So, try to adjust it to obtain the form same as the limit formula of trigonometric function.

If $h \to 0$, then $\dfrac{h}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \to 0$. It has proved that if $h$ approaches zero, then $\dfrac{h}{2}$ also approaches zero.

$=\,\,\,$ $\sec{x}\tan{x}$ $\times$ $\large \displaystyle \lim_{\frac{h}{2} \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

Take $z = \dfrac{h}{2}$ and express the limit of trigonometric function in terms of $z$.

$=\,\,\,$ $\sec{x}\tan{x}$ $\times$ $\large \displaystyle \lim_{z \,\to\, 0}{\normalsize \dfrac{\sin{z}}{z}}$

According to limit of sinx/x as x approaches 0 formula, the limit of the trigonometric function is equal to one.

$=\,\,\,$ $\sec{x}\tan{x} \times 1$

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

Therefore, it is proved that the derivative of secant function with respect to a variable is equal to the product of secant and tangent functions.

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