Which of the following functions is differentiable at `x = 0?`

Options

A. (a) `cos(|x|) + |x|`
B. (b) `cos (|x|) - |x|`
C. (c) `sin (|x|) + |x|`
D. (d) `sin (|x|) - |x|`

Answer

To determine which of the given functions is differentiable at \( x = 0 \), we will analyze each function step by step.

### Step 1: Understand the Functions
We have four functions:
1. \( a(x) = \cos |x| + |x| \)
2. \( b(x) = \cos |x| - |x| \)
3. \( c(x) = \sin |x| + |x| \)
4. \( d(x) = \sin |x| - |x| \)

### Step 2: Define the Absolute Value Function
The absolute value function \( |x| \) can be defined piecewise:
- \( |x| = x \) if \( x \geq 0 \)
- \( |x| = -x \) if \( x < 0 \)

### Step 3: Analyze Each Function

#### Function \( a(x) \)
For \( x \geq 0 \):
\[ a(x) = \cos x + x \]
For \( x < 0 \):
\[ a(x) = \cos(-x) - x = \cos x - (-x) = \cos x + x \]

**Derivative Calculation:**
- For \( x > 0 \): \( a'(x) = -\sin x + 1 \)
- For \( x < 0 \): \( a'(x) = -\sin x + 1 \)

At \( x = 0 \):
- Left-hand derivative: \( \lim_{h \to 0^-} a'(h) = -\sin(0) + 1 = 1 \)
- Right-hand derivative: \( \lim_{h \to 0^+} a'(h) = -\sin(0) + 1 = 1 \)

Since both derivatives are equal, \( a(x) \) is differentiable at \( x = 0 \).

#### Function \( b(x) \)
For \( x \geq 0 \):
\[ b(x) = \cos x - x \]
For \( x < 0 \):
\[ b(x) = \cos(-x) + x = \cos x + x \]

**Derivative Calculation:**
- For \( x > 0 \): \( b'(x) = -\sin x - 1 \)
- For \( x < 0 \): \( b'(x) = -\sin x + 1 \)

At \( x = 0 \):
- Left-hand derivative: \( \lim_{h \to 0^-} b'(h) = -\sin(0) + 1 = 1 \)
- Right-hand derivative: \( \lim_{h \to 0^+} b'(h) = -\sin(0) - 1 = -1 \)

Since the derivatives are not equal, \( b(x) \) is not differentiable at \( x = 0 \).

#### Function \( c(x) \)
For \( x \geq 0 \):
\[ c(x) = \sin x + x \]
For \( x < 0 \):
\[ c(x) = \sin(-x) + (-x) = -\sin x - x \]

**Derivative Calculation:**
- For \( x > 0 \): \( c'(x) = \cos x + 1 \)
- For \( x < 0 \): \( c'(x) = -\cos x - 1 \)

At \( x = 0 \):
- Left-hand derivative: \( \lim_{h \to 0^-} c'(h) = -\cos(0) - 1 = -2 \)
- Right-hand derivative: \( \lim_{h \to 0^+} c'(h) = \cos(0) + 1 = 2 \)

Since the derivatives are not equal, \( c(x) \) is not differentiable at \( x = 0 \).

#### Function \( d(x) \)
For \( x \geq 0 \):
\[ d(x) = \sin x - x \]
For \( x < 0 \):
\[ d(x) = \sin(-x) + x = -\sin x + x \]

**Derivative Calculation:**
- For \( x > 0 \): \( d'(x) = \cos x - 1 \)
- For \( x < 0 \): \( d'(x) = -\cos x + 1 \)

At \( x = 0 \):
- Left-hand derivative: \( \lim_{h \to 0^-} d'(h) = -\cos(0) + 1 = 0 \)
- Right-hand derivative: \( \lim_{h \to 0^+} d'(h) = \cos(0) - 1 = 0 \)

Since both derivatives are equal, \( d(x) \) is differentiable at \( x = 0 \).

### Conclusion
The only function that is differentiable at \( x = 0 \) is \( d(x) = \sin |x| - |x| \).