Question

Find the value of \( \frac{d(\tan x){dx} \)}

• A. \( \cot^2 x \)
• B. \( \sec^2 x \)
• C. \( \frac{1}{\tan x} \)
• D. \( \sec x \)

Hint:

Memorizing the derivatives of basic trigonometric functions (\( \sin x, \cos x, \tan x, \cot x, \sec x, \csc x \)) is essential for speed and accuracy in calculus problems. The derivative of \( \tan x \) is \( \sec^2 x \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question asks for the derivative of the trigonometric function \( \tan x \) with respect to x. This is a standard result in differential calculus.

Step 2: Key Formula or Approach:
We can find the derivative using the quotient rule, by first expressing \( \tan x \) as \( \frac{\sin x}{\cos x} \).
The quotient rule for differentiation states:
If \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).
We also need the standard derivatives:
\( \frac{d}{dx}(\sin x) = \cos x \)
\( \frac{d}{dx}(\cos x) = -\sin x \)

Step 3: Detailed Explanation:
Let \( y = \tan x = \frac{\sin x}{\cos x} \).
Here, \( u(x) = \sin x \) and \( v(x) = \cos x \).
First, find the derivatives of u(x) and v(x):
\( u'(x) = \frac{d}{dx}(\sin x) = \cos x \)
\( v'(x) = \frac{d}{dx}(\cos x) = -\sin x \)
Now, apply the quotient rule:
\[ \frac{dy}{dx} = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2} \] \[ \frac{dy}{dx} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \] Using the trigonometric identity \( \sin^2 x + \cos^2 x = 1 \), the numerator becomes 1.
\[ \frac{dy}{dx} = \frac{1}{\cos^2 x} \] Since \( \sec x = \frac{1}{\cos x} \), we have \( \sec^2 x = \frac{1}{\cos^2 x} \).
Therefore,
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]

Step 4: Final Answer:
The derivative of \( \tan x \) is \( \sec^2 x \). This is a standard derivative that is important to memorize.