Question

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

Answer 1

Hint: 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 \[h\left( x \right) = f\left( x \right) \cdot g\left( x \right)\].