What is the second derivative of \[f(x) = \sec x\]?

Hint: Here, the given question has a trigonometric function. We have to find the derivative or differentiated term of the function. We know the standard derivative of secant function and which is the first derivative. To find the second derivative we use product rule that is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\].

Complete step by step solution:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Given \[f(x) = \sec x\].
\[f(x) = \dfrac{1}{{\cos x}}\]
Or
\[f(x) = {\left( {\cos x} \right)^{ - 1}}\]
Now differentiate with respect to ‘x’, we have,
\[{f'}(x) = \dfrac{d}{{dx}}{\left( {\cos x} \right)^{ - 1}}\]
\[{f'}(x) = - {\left( {\cos x} \right)^{ - 1 - 1}}\dfrac{d}{{dx}}\left( {\cos x} \right)\]
\[{f'}(x) = - {\left( {\cos x} \right)^{ - 2}}\left( { - \sin x} \right)\]
\[{f'}(x) = \dfrac{{\sin x}}{{{{\left( {\cos x} \right)}^2}}}\]
\[{f'}(x) = \dfrac{{\sin x}}{{\cos x}}\dfrac{1}{{\cos x}}\]
\[{f'}(x) = \tan x.\sec x\]. This is the first derivative.
To find the second derivative we again differentiate the first derivative.
\[{f^{''}}(x) = \dfrac{d}{{dx}}\left( {\tan x.\sec x} \right)\]
Using the product rule of differentiation, we have,
\[{f^{''}}(x) = \tan x\dfrac{d}{{dx}}\left( {\sec x} \right) + \sec x\dfrac{d}{{dx}}\left( {\tan x} \right)\].
We know \[\dfrac{d}{{dx}}\left( {\sec x} \right) = \tan x.\sec x\] and \[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\],
\[{f^{''}}(x) = \tan x\left( {\sec x.\tan x} \right) + \sec x\left( {{{\sec }^2}x} \right)\]
\[{f^{''}}(x) = \sec x.{\tan ^2}x + {\sec ^3}x\]. This is the required result.
Thus the second derivative of \[f(x) = \sec x\] is \[\sec x.{\tan ^2}x + {\sec ^3}x\].

Additional information:
\[ \bullet \] Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as \[h'(x) = af'(x) + bg'(x)\]
\[ \bullet \] Quotient rule: The derivative of one function divided by other is found by quotient rule such as\[{\left[ {\dfrac{{f(x)}}{{g(x)}}} \right]'} = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left[ {g(x)} \right]}^2}}}\].
\[ \bullet \] Product rule: When a derivative of a product of two function is to be found, then we use product rule that is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\].
\[ \bullet \] Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is \[fog'({x_0}) = [(f'og)({x_0})]g'({x_0})\].

Note:
We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\]. The first obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.