Question

Let $ \phi_1(x) = e^{\sin x, \phi_2(x)} = e^{\phi_1(x)}, \ldots, \phi_{n+1}(x) = e^{\phi_n(x)}, \forall n \geq 1$. Then for any fixed $ n $, the expression $ \frac{d}{dx} \left\{ \phi_n(x) \right\} $ is:

• A. \( \phi_n(x) \cdot \phi_{n-1}(x) \)
• B. \( \phi_n(x) \cdot \phi_{n-1}(x) \cdot \ldots \cdot \phi_1(x) \cos x \)
• C. \( \phi_n(x) \cdot \phi_{n-1}(x) \cdot \ldots \cdot \varphi_1(x) \sin x \)
• D. \( \phi_n(x) \cdot \phi_{n-1}(x) \cdot \ldots \cdot \varphi_1(x) e^{\sin x} \)

Hint:

When differentiating composite functions, always apply the chain rule at each level. For such nested functions, you multiply each derivative step by step.

Answer 1

The Correct Option is B

Step 1: Recognize the function structure.
The functions \( \phi_n(x) \) are defined recursively: \[ \phi_1(x) = e^{\sin x}, \quad \phi_2(x) = e^{\phi_1(x)}, \quad \phi_3(x) = e^{\phi_2(x)}, \quad \ldots \] For any \( n \), \( \phi_n(x) = e^{\phi_{n-1}(x)} \).
Step 2: Apply the chain rule of differentiation.
To differentiate \( \phi_n(x) \), we use the chain rule. For example, to differentiate \( \phi_2(x) = e^{\phi_1(x)} \), we get: \[ \frac{d}{dx} \left( \phi_2(x) \right) = e^{\phi_1(x)} \cdot \frac{d}{dx} \left( \phi_1(x) \right) = \phi_2(x) \cdot \frac{d}{dx} \left( e^{\sin x} \right). \]
Step 3: Differentiate \( \phi_1(x) \).
We now differentiate \( \phi_1(x) = e^{\sin x} \). Using the chain rule again, we get: \[ \frac{d}{dx} \left( e^{\sin x} \right) = e^{\sin x} \cdot \cos x. \] Thus, \( \frac{d}{dx} \left( \phi_1(x) \right) = \phi_1(x) \cdot \cos x \).
Step 4: Generalize the differentiation for \( \phi_n(x) \).
Now, applying the chain rule recursively for \( \phi_n(x) \), we get: \[ \frac{d}{dx} \left\{ \phi_n(x) \right\} = \phi_n(x) \cdot \frac{d}{dx} \left\{ \phi_{n-1}(x) \right\} \cdot \ldots \cdot \frac{d}{dx} \left\{ \phi_1(x) \right\} \cdot \cos x. \] Thus, the derivative involves the product of all the terms up to \( \phi_n(x) \) and is multiplied by \( \cos x \).
Step 5:
Conclusion and simplification.

Therefore, the correct answer is the expression: \[ \frac{d}{dx} \left\{ \phi_n(x) \right\} = \phi_n(x) \cdot \phi_{n-1}(x) \cdot \ldots \cdot \phi_1(x) \cdot \cos x. \] This matches option (B).