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# Differentiate ${{\sec }^{-1}}x$ with respect to x using first principle.

Last updated date: 13th Jun 2024
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Hint: Here, the given function is inverse function so calculation will be tough but we can solve the question by using some standard results of trigonometric ratios and then we will substitute all the values in first principle formula to evaluate the derivative of ${{\sec }^{-1}}x$ with respect to x using first principle.

Before attempting this question let us see what is the first principle of derivative.
Suppose we have a real valued function f , the function defined by $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$ wherever the limit exist is defined to be the derivative of the function f at x and is denoted by $\dfrac{dy}{dx}$ or f ’ (x). this definition of derivative is also called the first principle of derivative thus, $\dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$ .
Now, before we start solving we see some identities which will help us in solving the question
We know that, $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$
Re - arranging the identity, we get
${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$
${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$
$\tan \theta =\sqrt{{{\sec }^{2}}\theta -1}$ …… ( i )
Let, $\sec \theta =x$
Taking ${{\sec }^{-1}}$ on both sides we get
$\theta ={{\sec }^{-1}}x$
Putting $\sec \theta =x$ in equation ( i ), we get
$\tan \theta =\sqrt{{{x}^{2}}-1}$
Taking ${{\tan }^{-1}}$ on both side we get
$\theta ={{\tan }^{-1}}\sqrt{{{x}^{2}}-1}$
Or ${{\sec }^{-1}}x={{\tan }^{-1}}\sqrt{{{x}^{2}}-1}$
Now , considering all the above equations we can now solve the derivative of ${{\sec }^{-1}}x$ w.r.t x,
Now, according to first principle
$\dfrac{d}{dx}({{\sec }^{-1}}x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{se{{c}^{-1}}(x+h)-{{\sec }^{-1}}(x)}{h}$
Putting ${{\sec }^{-1}}x={{\tan }^{-1}}\sqrt{{{x}^{2}}-1}$ , we get
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\sqrt{{{(x+h)}^{2}}-1}-{{\tan }^{-1}}\sqrt{({{x}^{2}}-1)}}{h}$
We know that, ${{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+A\cdot B} \right)$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\left( \dfrac{\sqrt{{{(x+h)}^{2}}-1}-\sqrt{({{x}^{2}}-1)}}{1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)}} \right)}{h}$
Multiplying numerator and denominator by $\dfrac{\sqrt{{{(x+h)}^{2}}-1}-\sqrt{({{x}^{2}}-1)}}{1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)}}$, we get
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\left( \dfrac{\sqrt{{{(x+h)}^{2}}-1}-\sqrt{({{x}^{2}}-1)}}{1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)}} \right)}{h\left( \dfrac{\sqrt{{{(x+h)}^{2}}-1}-\sqrt{({{x}^{2}}-1)}}{1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)}} \right)}\cdot \left( \dfrac{\sqrt{{{(x+h)}^{2}}-1}-\sqrt{({{x}^{2}}-1)}}{1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)}} \right)$
We know that, $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}x}{x}=1$, so
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\cdot \left( \dfrac{\sqrt{{{(x+h)}^{2}}-1}-\sqrt{({{x}^{2}}-1)}}{1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)}} \right)$
Multiplying numerator and denominator by $\sqrt{{{(x+h)}^{2}}-1}+\sqrt{({{x}^{2}}-1)}$, we get
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\cdot \left( \dfrac{\left( \sqrt{{{(x+h)}^{2}}-1}-\sqrt{({{x}^{2}}-1)} \right)\left( \sqrt{{{(x+h)}^{2}}-1}+\sqrt{({{x}^{2}}-1)} \right)}{\left( 1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)} \right)\left( \sqrt{{{(x+h)}^{2}}-1}+\sqrt{({{x}^{2}}-1)} \right)} \right)$
On solving using identity ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$ , we get
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\cdot \left( \dfrac{{{(x+h)}^{2}}-1-{{x}^{2}}+1}{\left( 1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)} \right)\left( \sqrt{{{(x+h)}^{2}}-1}+\sqrt{({{x}^{2}}-1)} \right)} \right)$
On solving we get
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\cdot \left( \dfrac{{{(x+h)}^{2}}-{{x}^{2}}}{\left( 1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)} \right)\left( \sqrt{{{(x+h)}^{2}}-1}+\sqrt{({{x}^{2}}-1)} \right)} \right)$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\cdot \left( \dfrac{h(2x+h)}{\left( 1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)} \right)\left( \sqrt{{{(x+h)}^{2}}-1}+\sqrt{({{x}^{2}}-1)} \right)} \right)$
On, solving we get
$=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{(2x+h)}{\left( 1+\sqrt{{{(x+h)}^{2}}-1}\cdot \sqrt{({{x}^{2}}-1)} \right)\left( \sqrt{{{(x+h)}^{2}}-1}+\sqrt{({{x}^{2}}-1)} \right)} \right)$…….( ii )
Now putting limits $h\to 0$in equation ( ii ), we get
$=\left( \dfrac{2x}{\left( 1+\sqrt{({{x}^{2}}-1)}\cdot \sqrt{({{x}^{2}}-1)} \right)\left( \sqrt{({{x}^{2}}-1)}+\sqrt{({{x}^{2}}-1)} \right)} \right)$
On solving,
$=\left( \dfrac{2x}{\left( 1+({{x}^{2}}-1) \right)\left( 2\sqrt{({{x}^{2}}-1)} \right)} \right)$
On simplifying, we get
$=\dfrac{1}{x\sqrt{({{x}^{2}}-1)}}$ , for $x\ge 1$ as ${{\sec }^{-1}}x={{\sec }^{-1}}x,x\ge 1$

Note: Solving differentials of inverse functions using first principle are tough so there might be chances of calculation error which can affect the solving of the question so one must avoid the errors and trigonometric identities must be known so that they can be used to simplify the terms.