Let `f(x)=sinx,g(x)=2x" and "h(x)=cosx.` If `phi(x)=["go"(fh)](x)," then "phi''((pi)/(4))` is equal to

Options

A. 4
B. 0
C. -4
D. none of these

Explanation

To solve the problem step by step, we start with the given functions and proceed to find the second derivative of the function \(\phi(x)\).

### Step 1: Define the Functions
We are given:
- \( f(x) = \sin x \)
- \( g(x) = 2x \)
- \( h(x) = \cos x \)

### Step 2: Find \( f(h(x)) \)
We need to find \( f(h(x)) \):
\[ h(x) = \cos x \implies f(h(x)) = f(\cos x) = \sin(\cos x) \]

### Step 3: Find \( g(f(h(x))) \)
Now we compute \( g(f(h(x))) \):
\[ g(f(h(x))) = g(\sin(\cos x)) = 2\sin(\cos x) \]

### Step 4: Define \(\phi(x)\)
Thus, we can define \(\phi(x)\):
\[ \phi(x) = g(f(h(x))) = 2\sin(\cos x) \]

### Step 5: Differentiate \(\phi(x)\) to Find \(\phi'(x)\)
Now we differentiate \(\phi(x)\):
\[ \phi'(x) = \frac{d}{dx}[2\sin(\cos x)] = 2\cos(\cos x) \cdot (-\sin x) = -2\sin x \cos(\cos x) \]

### Step 6: Differentiate Again to Find \(\phi''(x)\)
Now we differentiate \(\phi'(x)\) to find \(\phi''(x)\):
\[ \phi''(x) = \frac{d}{dx}[-2\sin x \cos(\cos x)] \]
Using the product rule:
\[ \phi''(x) = -2[\cos x \cos(\cos x) + \sin x \sin(\cos x)(-\sin x)] \]
This simplifies to:
\[ \phi''(x) = -2[\cos x \cos(\cos x) - \sin^2 x \sin(\cos x)] \]

### Step 7: Evaluate \(\phi''\left(\frac{\pi}{4}\right)\)
Now we evaluate \(\phi''\left(\frac{\pi}{4}\right)\):
1. Calculate \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)
2. Calculate \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)
3. Calculate \(\cos\left(\cos\left(\frac{\pi}{4}\right)\right) = \cos\left(\frac{1}{\sqrt{2}}\right)\)
4. Calculate \(\sin\left(\cos\left(\frac{\pi}{4}\right)\right) = \sin\left(\frac{1}{\sqrt{2}}\right)\)

Substituting these values into the expression for \(\phi''\left(\frac{\pi}{4}\right)\):
\[ \phi''\left(\frac{\pi}{4}\right) = -2\left[\frac{1}{\sqrt{2}} \cos\left(\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}}\right)^2 \sin\left(\frac{1}{\sqrt{2}}\right)\right] \]
This simplifies to:
\[ \phi''\left(\frac{\pi}{4}\right) = -2\left[\frac{1}{\sqrt{2}} \cos\left(\frac{1}{\sqrt{2}}\right) - \frac{1}{2} \sin\left(\frac{1}{\sqrt{2}}\right)\right] \]

### Final Result
After evaluating the above expression, we find that:
\[ \phi''\left(\frac{\pi}{4}\right) = -4 \]