Math Doubts

Derivative of secx formula Proof

The derivative of secant function with respect to a variable is equal to the product of secant and tangent. If $x$ represents a variable, then the secant function is written as $\sec{x}$. The differentiation of the $\sec{x}$ with respect to $x$ is equal to product of $\sec{x}$ and $\tan{x}$. The derivative of secant function is mathematically derived by first principle.

Express Differentiation of function in Limit form

The derivative of function in terms of $x$ is written in the following limiting operation form by the definition of the derivative.

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

If $f{(x)} = \sec{x}$, then $f{(x+h)} = \sec{(x+h)}$. The proof of differentiation of $\sec{x}$ function with respect to $x$ can be started from 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

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

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

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

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

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.

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

The limit of product of two functions can be valuated by finding the product of their limits by the product rule of limits.

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

The factor $2$ in the numerator of the second function is multiplying the other factor and it divides the denominator. So, it can be shifted 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

Firstly, find the limit of the first trigonometric function by the direct substitution method.

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

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

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

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

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

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

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

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

The limit of second trigonometric function is almost similar to the limit of $\dfrac{\sin{x}}{x}$ as $x$ approaches $0$ formula but the input is slightly different. So, we have to set the input same as the limit rule.

If $h \to 0$, then $\dfrac{h}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \to 0$. It is 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.

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

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



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