`d/(dx)[cos(1-x^2)^2]=`

Options

A. `-2x(1-x^2) sin (1-x^2)^2`
B. `-4x(1-x^2) sin (1-x^2)^2`
C. `4x (1-x^2) sin (1-x^2)^2`
D. `-2(1-x^2) sin (1-x^2)^2`

Answer

To solve the problem \( \frac{d}{dx} \left[ \cos(1 - x^2)^2 \right] \), we will use the chain rule and the product rule of differentiation. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
Let \( y = \cos(z) \) where \( z = (1 - x^2)^2 \).

### Step 2: Differentiate using the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} \]

### Step 3: Differentiate \( y \) with respect to \( z \)
Now, differentiate \( y = \cos(z) \):
\[ \frac{dy}{dz} = -\sin(z) \]

### Step 4: Differentiate \( z \) with respect to \( x \)
Next, we need to differentiate \( z = (1 - x^2)^2 \):
Using the chain rule again:
\[ \frac{dz}{dx} = 2(1 - x^2) \cdot \frac{d}{dx}(1 - x^2) \]
Now, differentiate \( 1 - x^2 \):
\[ \frac{d}{dx}(1 - x^2) = -2x \]
So,
\[ \frac{dz}{dx} = 2(1 - x^2)(-2x) = -4x(1 - x^2) \]

### Step 5: Combine the results
Now we can substitute \( \frac{dy}{dz} \) and \( \frac{dz}{dx} \) back into the chain rule:
\[ \frac{dy}{dx} = -\sin(z) \cdot (-4x(1 - x^2)) = 4x(1 - x^2) \sin(z) \]

### Step 6: Substitute back for \( z \)
Recall that \( z = (1 - x^2)^2 \):
\[ \frac{dy}{dx} = 4x(1 - x^2) \sin((1 - x^2)^2) \]

### Final Answer
Thus, the derivative is:
\[ \frac{d}{dx} \left[ \cos(1 - x^2)^2 \right] = 4x(1 - x^2) \sin((1 - x^2)^2) \]