If f(x) = |x - 2| and g(x) = f[f(x)], then g'(x) = ......... for x 20

Q: Wait, why can't I just use the chain rule on the absolute value function directly? Like, the derivative of $|u|$ is $u' \cdot \text{sgn}(u)$?

You absolutely *can* use the chain rule, but it is a trap! The signum function $\text{sgn}(u)$ depends entirely on the value of $u$. If you don't evaluate the sign of $u$ first, you will get stuck in algebraic ambiguity. By simplifying the function based on the domain $x > 20$ first, you avoid the signum function entirely. In JEE Advanced, always simplify the function *before* you differentiate to save time and avoid errors.

Q: Does the $x > 20$ condition really matter? Can't I just find the derivative for all $x$?

That is a brilliant question! If you were asked for the derivative for all $x$, you would have to define $g(x)$ as a piecewise function because the behavior changes at the 'kinks' (where the argument of the modulus is zero). But here, the problem specifically restricts us to $x > 20$. This is a gift! It allows us to ignore the complex piecewise behavior and focus on the linear region. Always respect the domain constraints; they are there to simplify your life.

Q: Why did the vertex shift from $x=2$ to $x=4$?

Think of the composition $f(f(x))$ as a transformation. The inner function $f(x) = |x-2|$ shifts the graph to the right by 2. When you apply $f$ again, you are essentially taking the output of that first shift and shifting it again. When you trace the algebra $f(f(x)) = | |x-2| - 2 |$, you are effectively creating a new 'V' shape whose vertex is determined by where the inner expression equals zero. Setting $|x-2| - 2 = 0$ gives $|x-2| = 2$, which leads to $x-2 = 2$ or $x-2 = -2$. Thus, the vertices are at $x=4$ and $x=0$. Since we are in the $x > 20$ region, we are far to the right of these vertices!

Question

If $f(x) = |x-2|$ and $g(x) = f[f(x)]$, then $g'(x) = \dots\dots\dots$ for $x > 20$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a function. When you see $f(x) = |x-2|$, I want you to immediately visualize a sharp, elegant 'V' shape.

The vertex of this 'V' sits firmly at $x = 2$. It is a simple, beautiful geometric object. But the problem asks us to go deeper. We are looking at a composite function, $g(x) = f(f(x))$, and we are restricted to a very specific, high-value domain: $x > 20$.

### The Power of the Domain Constraint

Many students see the absolute value and immediately panic, thinking about piecewise definitions and case-by-case analysis. But look at the domain! We are told $x > 20$. This is not just a number; it is a safety zone.

Let us analyze the inner function $f(x) = |x-2|$. If $x > 20$, then $x-2$ is clearly greater than $18$. Since $18$ is positive, the expression $x-2$ is strictly positive.

By the very definition of the modulus, $|a| = a$ when $a > 0$. Therefore, in our region of interest, the modulus brackets are essentially invisible. We can write:

$$f(x) = x - 2 \quad 	ext{for} \quad x > 20$$

This is our first victory. We have tamed the modulus. We have turned a potentially complex piecewise function into a simple, linear equation.

### Constructing the Composite Function

Now, let us build the composite function $g(x) = f(f(x))$. This is where the magic happens. We take our definition of $f(x)$ and feed it into itself.

$$g(x) = f(f(x)) = |f(x) - 2|$$

Substitute our simplified $f(x) = x-2$ into this expression:

$$g(x) = |(x-2) - 2|$$

Look at the algebra inside the modulus. It is just a simple subtraction of constants. We have $-2 - 2$, which gives us $-4$. So, our composite function simplifies to:

$$g(x) = |x - 4|$$

### The Final Simplification

We are almost there, but we must be careful. We have $g(x) = |x-4|$. Do we need to worry about the modulus here? Again, look at our domain. We are still operating under the condition $x > 20$.

If $x > 20$, then $x - 4$ is definitely greater than $16$. Since $16$ is positive, the expression $x-4$ is guaranteed to be positive. Once again, the modulus brackets vanish, leaving us with a beautiful, linear function:

$$g(x) = x - 4$$

### The Final Differentiation

Now, the calculus becomes trivial. We are asked to find $g'(x)$. Since we have established that $g(x) = x - 4$ for all $x > 20$, we simply differentiate this linear expression with respect to $x$:

$$g'(x) = \frac{d}{dx}(x - 4)$$

Using the linearity of the derivative, we have:

$$g'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(4)$$

We know that the derivative of $x$ is $1$, and the derivative of any constant like $4$ is $0$. Thus:

$$g'(x) = 1 - 0 = \mathbf{1}$$

### The Takeaway

Think about what we just did. We took a nested, potentially confusing function and, by respecting the domain, we stripped away the complexity layer by layer. Geometrically, this makes perfect sense.

For $x > 20$, the function $g(x)$ is just a straight line with a slope of $1$. The derivative, which represents the slope, must be constant.

Never let the notation intimidate you. Whether it is $|x-2|$ or a complex composition, always look for the domain, simplify the expression, and let the calculus follow naturally. You have mastered the logic. Keep this clarity, and you will conquer any problem the JEE throws at you.

Animated Solution for Mathematics - Differentiation: If $f (x) = ∣ x - 2∣$ and $g (x) = f [f (x)]$ , then $g^{'} (x) = \dots\dots\dots$ for $x > 20$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Function $f (x)$

Simplifying $f (x)$ for $x > 20$

Setting up $g (x) = f (f (x))$

Simplifying Inside the Modulus

Removing the Final Modulus

Differentiating $g (x)$

Final Conclusion

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Power of the Domain Constraint

Constructing the Composite Function

The Final Simplification

The Final Differentiation

The Takeaway

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∣x−2∣ and g(x)=f[f(x)], then g′(x)=……… for x>20.

Enter Numerical Value:

Visualized Solution

Visualizing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>20

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)=f(f(x))