How do you differentiate \[\dfrac{x}{{\cos x}}\]?

Hint: In this question, we will differentiate the given expression by using the division rule of differentiation. Use the formula of derivatives and then simplify the answer by using trigonometric ratios to get the final answer.

Complete step by step answer:
Here we have to differentiate \[\dfrac{x}{{\cos x}}\].
Now differentiating \[\dfrac{x}{{\cos x}}\] w.r.t \[x\], we have
\[\dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = ?\]
We know that if \[f\left( x \right)\] and \[g\left( x \right)\] are functions of \[x\] then derivative of \[\dfrac{{f\left( x \right)}}{{g\left( x \right)}}\] with respective of \[x\] is given by \[\dfrac{{g\left( x \right)\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} - f\left( x \right)\dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}}}}{{{{\left( {g\left( x \right)} \right)}^2}}}\].
So, we have \[\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} - f\left( x \right)\dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}}}}{{{{\left( {g\left( x \right)} \right)}^2}}}\].
By using the above formula, we get
\[
   \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = \dfrac{{\cos x\dfrac{{d\left( x \right)}}{{dx}} - x\dfrac{{d\left( {\cos x} \right)}}{{dx}}}}{{{{\left( {\cos x} \right)}^2}}} \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = \dfrac{{\cos x\left( 1 \right) - x\dfrac{{d\left( {\cos x} \right)}}{{dx}}}}{{{{\cos }^2}x}} \\
\]
We know that \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]. By substituting this value, we have
\[
   \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = \dfrac{{\cos x - x\left( { - \sin x} \right)}}{{{{\cos }^2}x}} \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = \dfrac{{\cos x + x\left( {\sin x} \right)}}{{{{\cos }^2}x}} \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = \dfrac{{\cos x + x\sin x}}{{{{\cos }^2}x}} \\
\]
Splitting the terms on right-hand side, we have
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = \dfrac{{\cos x}}{{{{\cos }^2}x}} + \dfrac{{x\sin x}}{{{{\cos }^2}x}}\]
Cancelling the common terms, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = \dfrac{1}{{\cos x}} + \dfrac{{x\sin x}}{{\cos x}}\dfrac{1}{{\cos x}}\]
We know that \[\dfrac{1}{{\cos x}} = \sec x\] and \[\dfrac{{\sin x}}{{\cos x}} = \tan x\]. Substituting this value, we have
\[\therefore \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right) = \sec x + x\tan x\sec x\]

Thus, the derivative of \[\dfrac{x}{{\cos x}}\] is \[\sec x + x\tan x\sec x\].

Note: In mathematics, division rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Let \[f\left( x \right)\] and \[g\left( x \right)\] are functions of \[x\] then derivative of \[\dfrac{{f\left( x \right)}}{{g\left( x \right)}}\] with respective of \[x\] is given by \[\dfrac{{g\left( x \right)\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} - f\left( x \right)\dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}}}}{{{{\left( {g\left( x \right)} \right)}^2}}}\].