Question

The derivative of $\tan x - x$ with respect to $x$ is
A. $1 - {\tan ^2}x$
B. $\tan x$
C. $- {\tan ^2}x$
D. ${\tan ^2}x$

Answer 1

Derivative is nothing but a displacement or rate of change of any function.
If f(x) is any function with an independent variable x then its derivative can be written as $y = f'(x)$ where y is the dependent variable (dependent on x). We have many standard derivative formulas to find the derivative that contains trigonometric functions. Using those formulas we can easily solve this problem.
Formula: Some formula that we need to know:
$\dfrac{d}{{dx}}(x) = 1$
$\dfrac{d}{{dx}}\tan (x) = {\sec ^2}x$
${\tan ^2}x + {\sec ^2}x = 1$

Complete step by step answer:
Let us name the given function as $f(x) = \tan (x) - x$ . We aim to find the derivative of the function $f(x)$ . Derivative of a function $f(x)$ can be denoted as $f'(x)$ .
Now let’s find the derivative of the given function.
$f(x) = \tan (x) - x$
Now let’s differentiate the above function concerning $x$ by using the formula $\dfrac{d}{{dx}}(x) = 1$ and$\dfrac{d}{{dx}}\tan (x) = {\sec ^2}x$ we get
$f'(x) = {\sec ^2}x - 1$
From the trigonometric identities we have ${\tan ^2}x + {\sec ^2}x = 1$ . Let’s rewrite this formula for our convenience.
${\tan ^2}x + {\sec ^2}x = 1$ $\Rightarrow {\sec ^2}x - 1 = {\tan ^2}x$
Now let’s use this formula to simplify the derivative that we found.
$f'(x) = {\sec ^2}x - 1 = {\tan ^2}x$
Thus, we get $f'(x) = {\tan ^2}x$
That is the derivative of $\tan x - x$ with respect $x$ is ${\tan ^2}x$ .

So, the correct answer is “Option D”.

Note: As we know that there are many standard formulas for the derivatives of trigonometric function by using those formulas, we can find the derivative but to simplify the derivative we can use trigonometric identities.
If the options don’t have the derivatives directly, we can try simplifying the derivative to see whether the simplification matches any of the options given.