# Proof of Derivative Rule of Inverse Secant function

Let $x$ denotes a variable, which represents a real number. In inverse trigonometry, the inverse secant function is written as $\sec^{-1}{(x)}$ or $\operatorname{arcsec}{(x)}$ mathematically. The derivative or the differentiation of the inverse sec function with respect to $x$ is written in differential calculus in two forms as follows.

$(1) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\sec^{-1}{(x)}\Big)}$

$(2) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\operatorname{arcsec}{(x)}\Big)}$

The first principle of differentiation is used to derive the derivative of the inverse secant in differential calculus, where the differentiation of the inverse secant function is used as a formula. So, let us learn the proof of the formula for the differentiation of the inverse secant function.

### Derivative of Secant function in Limit form

The derivative of inverse secant function with respect to $x$ is written in limit form from the principle definition of the derivative.

$\dfrac{d}{dx}{\, (\sec^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\sec^{-1}{(x+\Delta x)}-\sec^{-1}{x}}{\Delta x}}$

In this case, the differential element $\Delta x$ can be written simply as $h$, if we consider $\Delta x = h$.

$\implies$ $\dfrac{d}{dx}{\, (\sec^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sec^{-1}{(x+h)}-\sec^{-1}{x}}{h}}$

Now, the differentiation of $\operatorname{arcsec}{(x)}$ function with respect to $x$ can be derived from first principle in differential calculus.

### Evaluate the Limit by Direct Substitution

First of all, let us try to evaluate the functionality of expression by the direct substitution as $h$ approaches $0$.

$= \,\,\,$ $\dfrac{\sec^{-1}{(x+0)}-\sec^{-1}{x}}{0}$

$= \,\,\,$ $\dfrac{\sec^{-1}{x}-\sec^{-1}{x}}{0}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\sec^{-1}{x}}-\cancel{\sec^{-1}{x}}}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

It is an indeterminate form, which cannot be evaluated mathematically. So, it is not possible to find the limit of the function by the direct substitution. Therefore, we have to choose another mathematical system for evaluating it.

### Simplify the Inverse trigonometric function

The direct substitution method is failed to evaluate the differentiation of inverse secant function. So, comeback to the first step for deriving the derivative of inverse secant function in another method.

$\implies$ $\dfrac{d}{dx}{\,(\sec^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sec^{-1}{(x+h)}-\sec^{-1}{x}}{h}}$

In calculus, there is no limit rule in inverse secant function but we have limit rules in inverse sine and inverse tan functions. Actually, it is not possible to express the inverse secant function in terms of inverse tan function. So, we must try to convert each inverse secant function in terms of inverse sine function.

According to the inverse trigonometry, we convert each inverse secant function into inverse cosine function firstly and then each inverse cosine function can be converted into inverse sine function.

According to reciprocal inverse trigonometric identity, $\sec^{-1}{x} \,=\, \cos^{-1}{\Big(\dfrac{1}{x}\Big)}$

$\implies$ $\dfrac{d}{dx}{\,(\sec^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos^{-1}{\Big(\dfrac{1}{x+h}\Big)}-\cos^{-1}{\Big(\dfrac{1}{x}\Big)}}{h}}$

As per the constant property, $\sin^{-1}{x}+\cos^{-1}{x} \,=\, \dfrac{\pi}{2}$. Therefore, $\cos^{-1}{x} \,=\, \dfrac{\pi}{2}-\sin^{-1}{x}$

$\implies$ $\dfrac{d}{dx}{\, (\sec^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(\dfrac{\pi}{2}-\sin^{-1}{\Big(\dfrac{1}{x+h}\Big)}\Big)-\Big(\dfrac{\pi}{2}-\sin^{-1}{\Big(\dfrac{1}{x}\Big)}\Big)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{\pi}{2}-\sin^{-1}{\Big(\dfrac{1}{x+h}\Big)}-\dfrac{\pi}{2}+\sin^{-1}{\Big(\dfrac{1}{x}\Big)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{\pi}{2}-\dfrac{\pi}{2}-\sin^{-1}{\Big(\dfrac{1}{x+h}\Big)}+\sin^{-1}{\Big(\dfrac{1}{x}\Big)}}{h}}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cancel{\dfrac{\pi}{2}}-\cancel{\dfrac{\pi}{2}}+\sin^{-1}{\Big(\dfrac{1}{x}\Big)}-\sin^{-1}{\Big(\dfrac{1}{x+h}\Big)}}{h}}$

$\implies$ $\dfrac{d}{dx}{\, (\sec^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Big(\dfrac{1}{x}\Big)}-\sin^{-1}{\Big(\dfrac{1}{x+h}\Big)}}{h}}$

The inverse trigonometric expression represents the difference of inverse sine function and it can be simplified by the difference identity of inverse sin functions.

$\sin^{-1}{x}-\sin^{-1}{y}$ $\,=\,$ $\sin^{-1}{\Big(x\sqrt{1-y^2}-y\sqrt{1-x^2}\Big)}$

As per this rule, the inverse trigonometric expression in the numerator can be simplified.

$\implies$ $\dfrac{d}{dx}{\, (\sec^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\Big( \dfrac{1}{x}\Big)\sqrt{1-\Big(\dfrac{1}{x+h}\Big)^2}-\Big(\dfrac{1}{x+h}\Big)\sqrt{1-\Big(\dfrac{1}{x}\Big)^2}}\Bigg)}{h}}$

Now, simplify the expression in argument of the inverse sine function.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1}{x}\sqrt{1-\dfrac{1^2}{(x+h)^2}}-\dfrac{1}{x+h}\sqrt{1-\dfrac{1^2}{x^2}}}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1}{x}\sqrt{1-\dfrac{1}{(x+h)^2}}-\dfrac{1}{x+h}\sqrt{1-\dfrac{1}{x^2}}}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1}{x}\sqrt{\dfrac{(x+h)^2-1}{(x+h)^2}}-\dfrac{1}{x+h}\sqrt{\dfrac{x^2-1}{x^2}}}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1}{x} \times \dfrac{\sqrt{(x+h)^2-1}}{(x+h)}-\dfrac{1}{x+h} \times \dfrac{\sqrt{x^2-1}}{x}}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1 \times \sqrt{(x+h)^2-1}}{x \times (x+h)}-\dfrac{1 \times \sqrt{x^2-1}}{(x+h) \times x}}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}}{x(x+h)}-\dfrac{\sqrt{x^2-1}}{(x+h)x}}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}}{x(x+h)}-\dfrac{\sqrt{x^2-1}}{x(x+h)}}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}}{h}}$

The limit of rational expression is similar to the limit rule of sine function but the expression in the denominator must be same as the argument in the inverse sin function. So, we have to do some adjustments for transforming it into our required form.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}}{h}}$ $\times$ $1 \Bigg]$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}}{h}}$ $\times$ $\dfrac{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}\Bigg]$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}}$ $\times$ $\dfrac{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}{h}\Bigg]$

### Evaluate the Limits of the functions

As per the product rule of limits, the limit of product of functions can be evaluated by the product of their limits.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}}{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}{h}}$

The first multiplying function is almost same as the limit rule of inverse sin function but the input of the limiting operation must be argument in the inverse sin function or denominator. So, we have to evaluate the approaching value of the expression $\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}$ firstly.

$(1)\,\,\,$ If $h \,\to\, 0$, then $x+h \,\to\, x+0$. Therefore, $x+h \,\to\, x$

$(2)\,\,\,$ If $x+h \,\to\, x$, then $x \times (x+h) \,\to\, x \times x$. Therefore, $x(x+h) \,\to\, x^2$

$(3)\,\,\,$ We know that $x+h \,\to\, x$ and $x(x+h) \,\to\, x^2$, then $\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}$ $\,\to\,$ $\dfrac{\sqrt{(x)^2-1}-\sqrt{x^2-1}}{x^2}$. Now, $\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}$ $\,\to\,$ $\dfrac{\sqrt{x^2-1}-\sqrt{x^2-1}}{x^2}$, then $\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}$ $\,\to\,$ $\dfrac{0}{x^2}$. Therefore, $\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}$ $\,\to\,$ $0$

Let $u \,=\, \dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}$, then $u\,\to\,0$ as $h\,\to\,0$.

Now, express the first multiplying function in $u$ and keep the second factor as it is.

$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Big(u\Big)}}{\Big(u\Big)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{u}}{u}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}{h}}$

According to the limit rule of inverse sine function, the limit of quotient of inverse sine function $\sin^{-1}{u}$ by $u$ as $u$ tends to zero is equal to one.

$=\,\,\,$ $1$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)}\Bigg)}{\dfrac{h}{1}}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)} \times \dfrac{1}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(\sqrt{(x+h)^2-1}-\sqrt{x^2-1}\Big)}{x(x+h)} \times \dfrac{1}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(\sqrt{(x+h)^2-1}-\sqrt{x^2-1}\Big) \times 1}{x(x+h) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x(x+h)h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{hx(x+h)}}$

Try direct substitution method as $h$ approaches $0$ and you will observe that it gives us an indeterminate form. Hence, it is not recommendable to find the limit by direct substitution. So, we have to think about another approach. Actually, the function is in radical form. So, we can use rationalization method for evaluating it.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{hx(x+h)} \times 1\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{hx(x+h)}}$ $\times$ $\dfrac{\sqrt{(x+h)^2-1}+\sqrt{x^2-1}}{\sqrt{(x+h)^2-1}+\sqrt{x^2-1}}\Bigg)$

We can now multiply both fractional function. The product of the numerators of both fractional functions can be written as difference of squares of the terms.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(\sqrt{(x+h)^2-1}\Big)^2-\Big(\sqrt{x^2-1}\Big)^2}{hx(x+h) \times \Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(x+h)^2-1-(x^2-1)}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(x+h)^2-1-x^2+1}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(x+h)^2-x^2-1+1}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(x+h)^2-x^2-\cancel{1}+\cancel{1}}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(x+h)^2-x^2}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

The expression in the numerator can be factored by the difference of squares formula.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(x+h+x)(x+h-x)}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(x+x+h)(x-x+h)}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(2x+h)(\cancel{x}-\cancel{x}+h)}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(2x+h)(h)}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{h(2x+h)}{hx(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cancel{h}(2x+h)}{\cancel{h}x(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2x+h}{x(x+h)\Big(\sqrt{(x+h)^2-1}+\sqrt{x^2-1}\Big)}}$

Now, we can evaluate the limit by the direct substitution as $h$ approaches zero.

$=\,\,\,$ $\dfrac{2x+0}{x(x+0)\Big(\sqrt{(x+0)^2-1}+\sqrt{x^2-1}\Big)}$

$=\,\,\,$ $\dfrac{2x}{x(x)\Big(\sqrt{(x)^2-1}+\sqrt{x^2-1}\Big)}$

$=\,\,\,$ $\dfrac{2x}{x(x)\Big(\sqrt{x^2-1}+\sqrt{x^2-1}\Big)}$

$=\,\,\,$ $\dfrac{2x}{x(x)\Big(2\sqrt{x^2-1}\Big)}$

$=\,\,\,$ $\dfrac{2x}{2x(x)\sqrt{x^2-1}}$

$=\,\,\,$ $\dfrac{\cancel{2x}}{\cancel{2x}(x)\sqrt{x^2-1}}$

$=\,\,\,$ $\dfrac{1}{x\sqrt{x^2-1}}$

In this case, $x \ne 0$ and $x^2-1 > 0$, then $x^2 > 1$. Therefore, $x > \pm 1$. Hence, we take $x$ as $|x|$ for absolute value.

$=\,\,\,$ $\dfrac{1}{|x|\sqrt{x^2-1}}$

In this way, the differentiation of the inverse secant function is derived mathematically by the first principle in differential calculus.

$\therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, (\sec^{-1}{x})}$ $\,=\,$ $\dfrac{1}{|x|\sqrt{x^2-1}}$

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

###### Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

###### Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

###### Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.