If `(d)/(dx)(tanx+(2)/(3)tan^(3)x+(1)/(5)tan^(5)x)=sec^nx," then "n=`

Options

A. 7
B. 6
C. 4
D. 5

Answer

To solve the problem, we need to differentiate the expression \( \tan x + \frac{2}{3} \tan^3 x + \frac{1}{5} \tan^5 x \) and set it equal to \( \sec^n x \). We will find the value of \( n \).

### Step-by-Step Solution:

1. **Differentiate each term separately**:
\[ \frac{d}{dx} \left( \tan x + \frac{2}{3} \tan^3 x + \frac{1}{5} \tan^5 x \right) = \frac{d}{dx} (\tan x) + \frac{2}{3} \frac{d}{dx} (\tan^3 x) + \frac{1}{5} \frac{d}{dx} (\tan^5 x) \]

2. **Apply the derivative of \( \tan x \)**:
\[ \frac{d}{dx} (\tan x) = \sec^2 x \]

3. **Use the chain rule for \( \tan^3 x \) and \( \tan^5 x \)**:
- For \( \tan^3 x \):
\[ \frac{d}{dx} (\tan^3 x) = 3 \tan^2 x \cdot \sec^2 x \]
- For \( \tan^5 x \):
\[ \frac{d}{dx} (\tan^5 x) = 5 \tan^4 x \cdot \sec^2 x \]

4. **Substituting back into the differentiation**:
\[ \sec^2 x + \frac{2}{3} \cdot 3 \tan^2 x \sec^2 x + \frac{1}{5} \cdot 5 \tan^4 x \sec^2 x \]
Simplifying this gives:
\[ \sec^2 x + 2 \tan^2 x \sec^2 x + \tan^4 x \sec^2 x \]

5. **Factor out \( \sec^2 x \)**:
\[ \sec^2 x \left( 1 + 2 \tan^2 x + \tan^4 x \right) \]

6. **Recognize the expression in the parentheses**:
The expression \( 1 + 2 \tan^2 x + \tan^4 x \) can be rewritten as:
\[ (1 + \tan^2 x)^2 \]
Therefore, we have:
\[ \sec^2 x (1 + \tan^2 x)^2 \]

7. **Using the identity \( 1 + \tan^2 x = \sec^2 x \)**:
\[ \sec^2 x (\sec^2 x)^2 = \sec^2 x \cdot \sec^4 x = \sec^6 x \]

8. **Setting the result equal to \( \sec^n x \)**:
\[ \sec^6 x = \sec^n x \]
This implies:
\[ n = 6 \]

### Final Answer:
\[ n = 6 \]