Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How do you differentiate $\dfrac{x}{{\cos x}}$?

Last updated date: 17th Sep 2024
Total views: 402k
Views today: 4.02k
Verified
402k+ views
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.

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 $\dfrac{{g\left( x \right)\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} - f\left( x \right)\dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}}}}{{{{\left( {g\left( x \right)} \right)}^2}}}$.