Question

Find the derivative of $y = \ln \left( {\sec x} \right)$.

Answer 1

In the given question, we have to find the derivative of $y = \ln \left( {\sec x} \right)$. For finding the derivative of this function we can use the chain rule which is given by, if $h\left( x \right) = f\left( {g\left( x \right)} \right)$ then $h'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)$ .

Complete step by step solution:
Here we have to find the derivative of the function $y = \ln \left( {\sec x} \right)$. To do so we can apply the Chain rule. The Chain rule states that for a function defined as
Now by using the chain rule which is given by if $h\left( x \right) = f\left( {g\left( x \right)} \right)$ then $h'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)$ .
We will get the function as $h\left( x \right) = f\left( {g\left( x \right)} \right)$, the derivative of $h\left( x \right)$ is given by $h'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)$. Comparing this form with the given function we observe, $g(x) = \sec x$, $f\left( {g(x)} \right) = \ln \left( {\sec x} \right)$ and $h\left( x \right) = y$.
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\ln \left( {\sec x} \right)} \right)}}{{dx}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\ln \left( {\sec x} \right)} \right)}}{{dx}} \cdot \dfrac{{d\left( {\sec x} \right)}}{{dx}}$ [ Using Chain rule.]
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sec x}} \cdot \sec x\tan x$
Simplifying further, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \tan x$
Hence the derivative of the given function is $\tan x$.

Note: Note that the Chain rule of differentiation is used when we have to differentiate composition of functions, like $h\left( x \right) = f\left( {g\left( x \right)} \right)$. However, the product rule of differentiation is used when we have to differentiate the product of two functions, like $h\left( x \right) = f\left( x \right) \cdot g\left( x \right)$.