Let `f (x)=sin x,g(x) = x^2, h(x) = log_e x. `If F(x) = (hogof) x `then F"(x) is equal to

Options

A. `2 cosec^3 x`
B. ` 2 cot x^2 - 4x^2 cosec^2 x^2`
C. `2x cot x^2`
D. `-2 cosec^2 x `

Answer

To solve the problem step by step, we need to find the second derivative \( F''(x) \) of the function \( F(x) = h(g(f(x))) \) where:

- \( f(x) = \sin x \)
- \( g(x) = x^2 \)
- \( h(x) = \log_e x \)

### Step 1: Find \( F(x) \)

First, we need to express \( F(x) \) in terms of \( x \):

1. Start with \( f(x) = \sin x \).
2. Then, apply \( g \) to \( f(x) \):
\[ g(f(x)) = g(\sin x) = (\sin x)^2 = \sin^2 x \]
3. Finally, apply \( h \) to \( g(f(x)) \):
\[ F(x) = h(g(f(x))) = h(\sin^2 x) = \log_e(\sin^2 x) \]

So, we have:
\[ F(x) = \log_e(\sin^2 x) \]

### Step 2: Differentiate \( F(x) \) to find \( F'(x) \)

Now, we will differentiate \( F(x) \):

Using the chain rule, the derivative of \( \log_e(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).

1. Differentiate \( F(x) \):
\[ F'(x) = \frac{1}{\sin^2 x} \cdot \frac{d}{dx}(\sin^2 x) \]
2. Now, differentiate \( \sin^2 x \) using the chain rule:
\[ \frac{d}{dx}(\sin^2 x) = 2\sin x \cdot \cos x \]
3. Substitute this back into the expression for \( F'(x) \):
\[ F'(x) = \frac{1}{\sin^2 x} \cdot (2\sin x \cdot \cos x) = \frac{2\cos x}{\sin x} = 2 \cot x \]

### Step 3: Differentiate \( F'(x) \) to find \( F''(x) \)

Now, we will differentiate \( F'(x) \) to find \( F''(x) \):

1. Differentiate \( F'(x) = 2 \cot x \):
\[ F''(x) = 2 \cdot \frac{d}{dx}(\cot x) \]
2. The derivative of \( \cot x \) is \( -\csc^2 x \):
\[ F''(x) = 2 \cdot (-\csc^2 x) = -2 \csc^2 x \]

### Final Result

Thus, the second derivative \( F''(x) \) is:
\[ F''(x) = -2 \csc^2 x \]