Question

What is the derivative of $\text{y}=\text{arcsec}\left( \text{x} \right)$?

Answer 1

To find the derivative of $\text{y}=\text{arcsec}\left( \text{x} \right)$ we will use the implicit differentiation. Firstly we will take the secant term on the left side and then differentiate it using implicit differentiation. Then we will simplify it and by using our original value and some trigonometric identities get our answer in terms of $x$ and hence get our desired answer.

Complete step by step solution:
To find the derivative of $\text{y}=\text{arcsec}\left( \text{x} \right)$ we will firstly write the $arc\sec \left( x \right)$ in its simplified form.
$\begin{aligned} & \Rightarrow y=arc\sec \left( x \right) \\\ & \therefore y={{\sec }^{-1}}\left( x \right) \\\ \end{aligned}$
Take the secant term on the left side we get,
$\sec y=x$…..$\left( 1 \right)$
Using Implicit differentiation on above value we get,
$\dfrac{d\left( \sec y \right)}{dx}=\dfrac{dx}{dx}$
We know differentiation of $\sec \left( y \right)$ is $\sec y\tan y$ so,
$\sec y\tan y\dfrac{dy}{dx}=1$
On simplifying we get,
$\dfrac{dy}{dx}=\dfrac{1}{\sec y\tan y}$
Substituting value from equation (1) in above we get,
$\dfrac{dy}{dx}=\dfrac{1}{x\tan y}$……$\left( 2 \right)$
Now, to find the value of $\tan y$ in terms of $x$ we will use the below identity.
${{\sec }^{2}}y=1+{{\tan }^{2}}y$
Substituting value from equation (1) in above equation we get,
$\begin{aligned} & \Rightarrow {{x}^{2}}=1+{{\tan }^{2}}y \\\ & \Rightarrow {{x}^{2}}-1={{\tan }^{2}}y \\\ & \Rightarrow \pm \sqrt{{{x}^{2}}-1}=\tan y \\\ \end{aligned}$
Substituting above value in equation (2) we get,
$\dfrac{dy}{dx}=\dfrac{1}{\pm x\sqrt{{{x}^{2}}-1}}$
Further we can say that if $x\ge 1$ or $x\le -1$ we will get our denominator term as positive in both case so we can rewrite the answer as:
$\dfrac{dy}{dx}=\dfrac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}$
Hence, derivative of $\text{y}=\text{arcsec}\left( \text{x} \right)$ is $\dfrac{1}{\left| x \right|\sqrt{1-{{x}^{2}}}}$

Note: The term Implicit differentiation used in the solution is done when some function having variable $x,y$ when differentiated doesn’t lead directly to the answer. In this method $\dfrac{dy}{dx}$ terms are collected in one side after differentiation and then we solve the equation further to get the derivative. Differentiation of Trigonometry functions is different from usual differentiation; this concept is very vast and useful. We start by finding the derivative of sine and cosine terms and then by using it obtain the derivatives of the remaining four trigonometric functions.