Derivative of sec x with respect to x

Formula

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

Proof

According to trigonometry, $x$ is an angle of a right angled triangle and the secant of the angle $x$ is written as $\sec x$. The differentiation of $\sec x$ with respect to $x$ can be derived by the differentiation rule in limit form.

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

Take $f(x) = \sec x$, then $f(x) = \sec(x+h)$. Substitute them in the differentiation rule in limit form.

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

Step: 1

The cosine and secant both are reciprocal trigonometric functions. Hence, express each secant function as cosine function as per the reciprocal trigonometric identity of cosine and secant functions.

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

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

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

Step: 2

Use sum to product transformation identity to express the numerator in product form.

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

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

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

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

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

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

Split the function as two multiplying sub functions to use limit sinx/x rule when x tends to 0.

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

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

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

If $h \to 0$ then $\dfrac{h}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \to 0$. Change limit of first function but no need to change the limit of second multiplying function.

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

As per the limit $\dfrac{\sin x}{x}$ rule when $x$ tends to $0$, the value of ratio of $\sin \Big(\dfrac{h}{2}\Big)$ to $\dfrac{h}{2}$ is one when $\dfrac{h}{2}$ tends to $0$. Substitute $h$ equals to $0$ in second multiplying limits function.

$\,=\,$ $1 \times \dfrac{\sin \Bigg[ \dfrac{2x+0}{2} \Bigg]}{\cos(x+0) \cos x}$

$\,=\,$ $\dfrac{\sin \Bigg[ \dfrac{2x}{2} \Bigg]}{\cos x \times \cos x}$

$\,=\,$ $\require{cancel} \dfrac{\sin \Bigg[ \dfrac{\cancel{2}x}{\cancel{2}} \Bigg]}{\cos x \cos x}$

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

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

$\,=\,$ $\sec x \tan x$

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

Therefore, it is proved that the derivative of secant of angle $x$ with respect to $x$ is equal to the product of secant of $x$ and tangent of $x$.


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