If `(d)/(dx)(x-tanx+(1)/(3)tan^(3)x)=tan^(n)x," then "n=`

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to differentiate the expression \( x - \tan x + \frac{1}{3} \tan^3 x \) and set it equal to \( \tan^n x \). Let's go through the steps:

### Step 1: Differentiate the expression
We start with the expression:
\[ f(x) = x - \tan x + \frac{1}{3} \tan^3 x \]
Now, we differentiate \( f(x) \) with respect to \( x \):
\[ \frac{d}{dx}(f(x)) = \frac{d}{dx}(x) - \frac{d}{dx}(\tan x) + \frac{d}{dx}\left(\frac{1}{3} \tan^3 x\right) \]

### Step 2: Apply the derivatives
1. The derivative of \( x \) is \( 1 \).
2. The derivative of \( \tan x \) is \( \sec^2 x \).
3. For \( \frac{1}{3} \tan^3 x \), we use the chain rule:
\[ \frac{d}{dx}\left(\frac{1}{3} \tan^3 x\right) = \frac{1}{3} \cdot 3 \tan^2 x \cdot \frac{d}{dx}(\tan x) = \tan^2 x \cdot \sec^2 x \]

Putting it all together, we have:
\[ \frac{d}{dx}(f(x)) = 1 - \sec^2 x + \tan^2 x \sec^2 x \]

### Step 3: Simplify the expression
Now we can simplify the expression:
\[ \frac{d}{dx}(f(x)) = 1 - \sec^2 x + \tan^2 x \sec^2 x \]
We can factor out \( \tan^2 x \):
\[ = 1 - \sec^2 x + \tan^2 x \sec^2 x = 1 - \sec^2 x(1 - \tan^2 x) \]
Since \( \sec^2 x - 1 = \tan^2 x \), we can rewrite:
\[ = 1 - \sec^2 x \cdot \tan^2 x \]
This simplifies to:
\[ = -\tan^2 x + \tan^2 x \sec^2 x \]

### Step 4: Factor the expression
Now we can factor the expression:
\[ = \tan^2 x (\sec^2 x - 1) = \tan^2 x \tan^2 x = \tan^4 x \]

### Step 5: Set equal to \( \tan^n x \)
According to the problem, we have:
\[ \frac{d}{dx}(f(x)) = \tan^4 x \]
Thus, we can equate:
\[ \tan^4 x = \tan^n x \]
This implies:
\[ n = 4 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{4} \]