Question

How do you find the derivative of $y = \ln \left| {\sec x + \tan x} \right|$?

Answer 1

First of all, substitute $\sec x + \tan x$ to some variable. Then apply the chain rule of differentiation which is $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}$. Differentiate it step by step and use formulas $\dfrac{d}{{dx}}\sec x = \sec x\tan x$ and $\dfrac{d}{{dx}}\tan x = {\sec ^2}x$. Replace the variable with $\sec x + \tan x$ to get the answer.

Complete step-by-step solution:
According to the question, we have been given a function and we have to determine the derivative of it.
The given function is:
$\Rightarrow y = \ln \left| {\sec x + \tan x} \right|$
We know that logarithm is defined only for positive values so anything under logarithm must always be positive. Thus we can remove the modulus from the equation. So we have:
$\Rightarrow y = \ln \left( {\sec x + \tan x} \right)$
If we substitute $\sec x + \tan x = t$ in the equation, we will get:
$\Rightarrow y = \ln t$
Now differentiating the function with respect to $x$, we’ll get:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\ln t$
Here we will apply chain rule of differentiation. This rule is stated below:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}$
So applying this rule in our differentiation, we’ll get:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dt}}\ln t \times \dfrac{{dt}}{{dx}}$
We know that the differentiation of $\ln x$ is $\dfrac{1}{x}$, using this, we have:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{t} \times \dfrac{{dt}}{{dx}}$
Putting back the value of $t$, we’ll get:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {\sec x + \tan x} \right)}} \times \dfrac{d}{{dx}}\left( {\sec x + \tan x} \right)$
Differentiating it step by step, we’ll get:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {\sec x + \tan x} \right)}}\left( {\dfrac{d}{{dx}}\sec x + \dfrac{d}{{dx}}\tan x} \right)$
From the formulas of differentiation, we have:
$\Rightarrow \dfrac{d}{{dx}}\sec x = \sec x\tan x{\text{ and }}\dfrac{d}{{dx}}\tan x = {\sec ^2}x$
Putting these formulas, we’ll get:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {\sec x + \tan x} \right)}}\left( {\sec x\tan x + {{\sec }^2}x} \right)$
Simplifying it further, we’ll get:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {\sec x + \tan x} \right)}}\left( {\tan x + \sec x} \right)\sec x$
Cancelling $\left( {\tan x + \sec x} \right)$ from both numerator and denominator, we have:
$\Rightarrow \dfrac{{dy}}{{dx}} = \sec x$

$\sec x$ is the required answer.

Note: Whenever we have to differentiate a composite function, we always use the chain rule of differentiation after substitution. This makes a complex looking function simple from where we can differentiate step by step. For example, consider the given composite function:
$\Rightarrow y = f\left( {g\left( x \right)} \right)$
To differentiate this function, we’ll substitute $g\left( x \right) = t$, we will have:
$\Rightarrow y = f\left( t \right)$
Now we can apply chain rule of differentiation as shown below:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}f\left( t \right) \times \dfrac{{dt}}{{dx}}$
Now this differentiation is simple and we can do it step by step. After doing this, we can put back the value of $t$ to get the answer.