Which of the following functions $f(x)$ are differentiable at $x = 0$?

(A) $|x|$

(B) $|x - 1|$

(C) $|\sin x|$

(D) $|\cos x|$

(E) $x^2$

Choose the correct answer from the options given below:

A function is differentiable at a point if it has a well-defined derivative there. If a graph has a sharp corner or cusp at a point, it's not differentiable there.

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For $f(x) = |x|$:

At $x = 0$, the graph makes a sharp V-shape.

Left side $(x < 0)$: $|x| = -x$, so slope $= -1$

Right side $(x > 0)$: $|x| = x$, so slope $= +1$

Since $-1 \neq +1$, the slopes don't match.

Not differentiable at $x = 0$.

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For $f(x) = |x - 1|$:

The corner of $|x - 1|$ is at $x = 1$, not at $x = 0$.

When $x = 0$, the graph is on the smooth part where $x < 1$.

For $x < 1$: $|x - 1| = 1 - x$

At $x = 0$: $f(x) = 1 - x$, which gives $f'(x) = -1$

No corner exists at $x = 0$.

Differentiable at $x = 0$.

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For $f(x) = |\sin x|$:

$\sin(0) = 0$, and $\sin x$ crosses zero at $x = 0$.

This means $|\sin x|$ has a corner at $x = 0$.

Left side: $\sin x$ is negative, so $|\sin x| = -\sin x$, giving slope $= -\cos(0) = -1$

Right side: $\sin x$ is positive, so $|\sin x| = \sin x$, giving slope $= \cos(0) = +1$

Since $-1 \neq +1$, there's a sharp corner.

Not differentiable at $x = 0$.

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For $f(x) = |\cos x|$:

$\cos(0) = 1$ (positive)

The function $\cos x$ stays positive near $x = 0$ (it only becomes zero at $x = \pm\frac{\pi}{2}$).

Near $x = 0$: $|\cos x| = \cos x$

Therefore: $f'(x) = -\sin x$

$f'(0) = -\sin(0) = 0$

No corner exists at $x = 0$.

Differentiable at $x = 0$.

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For $f(x) = x^2$:

This is a polynomial, smooth everywhere.

$f'(x) = 2x$

$f'(0) = 0$

Differentiable at $x = 0$.

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Functions (B) $|x-1|$, (D) $|\cos x|$, and (E) $x^2$ are differentiable at $x = 0$.