Q.If $f(x) = |\cos x|$, then $f'\left(\dfrac{\pi}{4}\right) =$ __________.

Understanding Differentiability of Absolute Value

The function $y = x|x|$ looks deceptively simple, but the absolute value hides a piecewise structure. The key insight is that $|x|$ behaves differently on either side of zero: it equals $x$ when $x \geq 0$ and equals $-x$ when $x < 0$. This means our function has two different algebraic forms depending on the sign of $x$.

Rather than wrestling with the absolute value directly, we rewrite the function in its piecewise form, then differentiate the piece that applies to our domain of interest.

Step-by-Step Solution

1. Rewrite using the definition of absolute value

   Recall that $|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$

   Therefore:

2. Identify the relevant piece for $x < 0$

   Since we're asked to find $\frac{dy}{dx}$ specifically for $x < 0$, we work with the second piece:

3. Differentiate using the power rule

   For $x < 0$, we have a simple polynomial: …

Why This Problem Tests a Core Idea

The absolute value function $|x-1|$ is the classic example of a function that is continuous everywhere but fails to be differentiable at the point where its "kink" occurs — here, at $x=1$. The negative sign in front just flips the V-shape upside down; it doesn't change the nature of the corner.

Continuity asks: Does the graph have a break? Differentiability asks: Does the graph have a unique tangent line? At a sharp corner, the answer to the first is "no" and to the second is "no" — that's exactly what we'll verify.

Step-by-Step Reasoning

1. Rewrite the function without the absolute value

The expression $|x-1|$ behaves differently depending on whether $x-1$ is positive or negative:

• If $x \ge 1$, then $x-1 \ge 0$, so $|x-1| = x-1$.
• If $x < 1$, then $x-1 < 0$, so $|x-1| = -(x-1) = 1-x$.

Therefore, $f(x) = -|x-1|$ becomes:

So the function is two straight lines meeting at $x=1$: for $x<1$ it's the line $y = x-1$ (slope $+1$), and for $x \ge 1$ it's the line $y = 1-x$ (slope $-1$).

2. Check continuity at $x=1$

A function is continuous at $x=1$ if:

1. $f(1)$ exists,
2. $\lim_{x \to 1} f(x)$ exists,
3. They are equal.

Left-hand limit ($x \to 1^-$):

For $x<1$, $f(x) = x-1$. As $x$ approaches 1 from the left, $x-1 \to 0$. So:

Right-hand limit ($x \to 1^+$):

For $x \ge 1$, $f(x) = 1-x$. As $x$ approaches 1 from the right, $1-x \to 0$. So:

Function value:

At $x=1$, using the $x \ge 1$ piece: $f(1) = 1 - 1 = 0$.

Since both one-sided limits equal $f(1) = 0$, the function is continuous at $x=1$.

3. Check differentiability at $x=1$ …

The key to this problem is understanding what absolute value does to differentiability. The function $|x|$ has a V-shaped graph — it is continuous everywhere, but at $x=0$ the slope changes abruptly from $-1$ to $+1$. That sharp corner means the derivative does not exist at $x=0$, even though the function is perfectly continuous there. The same logic applies to $|x-1|$ at $x=1$.

When you add two such functions, the sum inherits the continuity of each piece. But at a point where either term has a corner, the sum may also have a corner — unless the slopes happen to cancel, which they do not here.

Let’s check each point carefully.

1. Continuity at $x=0$ Compute the left-hand limit, right-hand limit, and the function value. For $x < 0$: $|x| = -x$, $|x-1| = -(x-1) = 1-x$, so

As $x \to 0^-$, $f(x) \to 1$.

For $x > 0$ but $x < 1$: $|x| = x$, $|x-1| = 1-x$, so

As $x \to 0^+$, $f(x) \to 1$.

Also $f(0) = |0| + |0-1| = 0 + 1 = 1$.

Since left limit = right limit = function value, $f$ is continuous at $x=0$.

2. Differentiability at $x=0$ The left-hand derivative uses the expression for $x<0$: $f(x) = 1 - 2x$, so $f'(x) = -2$. Hence

The right-hand derivative uses the expression for $0 < x < 1$: $f(x) = 1$, so $f'(x) = 0$. Hence

Since $-2 \neq 0$, the left and right derivatives are different. Therefore $f$ is not differentiable at $x=0$.

Understanding Differentiability of Absolute Value Functions

The absolute value function $|x|$ itself has a sharp corner at $x = 0$, making it continuous but not differentiable there. However, when we multiply $x$ by $|x|$, we're creating something different. The key insight is to rewrite $f(x)$ in piecewise form and check whether the pieces "join smoothly" at the origin.

Start by recalling that $|x| = x$ when $x \geq 0$ and $|x| = -x$ when $x < 0$. This gives us:

Notice that both pieces are parabolas, one opening upward and one downward, meeting at the origin.

Checking Continuity at $x = 0$

1. Compute the left-hand limit:

2. Compute the right-hand limit:

3. Evaluate at the point:

Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$, the function is continuous at $x = 0$.

Checking Differentiability at $x = 0$

Differentiability requires that the derivative from the left equals the derivative from the right. We use the definition of the derivative:

4. Left-hand derivative: $f'_-(0) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} (-h) = 0$ …

The left-hand derivative at $x=0$ is

$\lim_{h\to0^-}\frac{|0+h|-|0|}{h}=\lim_{h\to0^-}\frac{-h}{h}=-1,$ …

The left-hand derivative is the slope of the left branch from $(0,1)$ to $(1,0)$: $\dfrac{0-1}{1-0} = -1$, so Statement 1 is true. The right branch rises from $(1,0)$ to $(2,1)$ with slope $\dfrac{1-0}{2-1} = +1$. Since the left and right derivatives differ ($-1 \neq +1$), the …