Let`f(x)=sinx,g(x)=x^(2)` and `h(x)=log_(e)x.`
If `F(x)=("hog of ")(x)," then "F''(x)` is equal to

Options

A. `2cosec^(3)x`
B. `2cotx^(2)-4x^(2)cosec^(2)x^(2)`
C. `2xcotx^(2)`
D. `-2cosec^(2)x`

Explanation

To find the second derivative \( F''(x) \) of the function \( F(x) = h(g(f(x))) \), where \( f(x) = \sin x \), \( g(x) = x^2 \), and \( h(x) = \log_e x \), we will follow these steps:

### Step 1: Define the Composition of Functions
We start by substituting the functions into \( F(x) \):
\[ F(x) = h(g(f(x))) = h(g(\sin x)) = h(\sin^2 x) \]
Since \( g(x) = x^2 \), we have:
\[ g(\sin x) = (\sin x)^2 = \sin^2 x \]
Now substituting into \( h(x) \):
\[ F(x) = h(\sin^2 x) = \log_e(\sin^2 x) \]

### Step 2: Simplify the Function
Using the logarithmic identity \( \log_e(a^b) = b \log_e(a) \), we can simplify \( F(x) \):
\[ F(x) = \log_e(\sin^2 x) = 2 \log_e(\sin x) \]

### Step 3: Find the First Derivative \( F'(x) \)
Now we differentiate \( F(x) \):
\[ F'(x) = 2 \cdot \frac{d}{dx}(\log_e(\sin x)) \]
Using the chain rule, we have:
\[ \frac{d}{dx}(\log_e(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x \]
Thus,
\[ F'(x) = 2 \cdot \cot x \]

### Step 4: Find the Second Derivative \( F''(x) \)
Next, we differentiate \( F'(x) \):
\[ F''(x) = \frac{d}{dx}(2 \cot x) \]
The derivative of \( \cot x \) is \( -\csc^2 x \), therefore:
\[ F''(x) = 2 \cdot (-\csc^2 x) = -2 \csc^2 x \]

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

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