Animated Solution for Mathematics - Differentiation: For $x\in R,f(x)=|\log 2-\sin x|$ and $g(x)=f(f(x))$, then

Welcome, future engineer! Today, we are going to peel back the layers of a composite function. Our objective is to find the derivative of $g(x)=f(f(x))$ at $x=0$, where $f(x)=|\log 2-\sin x|$.

Often, when students see a modulus operator, they panic. But I want you to see it differently; a modulus is not a barrier, but a signpost that tells us the function behaves differently depending on the region.

Before we apply the Chain Rule, we must unlock the modulus. Imagine you are standing on the graph of $y=\log 2-\sin x$.

We know that $\log 2\approx 0.693$. Near $x=0$, $\sin x$ is very close to $0$.

Since $0.693>0$, the expression $\log 2-\sin x$ is strictly positive in the neighborhood of $x=0$. This means the modulus bars are redundant, and we can write:

Now, let us look at $g(x)=f(f(x))$. Substituting our simplified $f(x)$, we get $g(x)=|\log 2-\sin (f(x))|$.

As $x\to 0$, $f(x)\to f(0)=\log 2-\sin (0)=\log 2$. The inner expression approaches $\log 2-\sin (\log 2)$.

Since $\log 2\approx 0.693$ and $\sin (0.693)\approx 0.638$, the difference is clearly positive. The second modulus also drops away, leaving us with the smooth function:

Now, we are ready for the calculus. We apply the Chain Rule to differentiate $g(x)$:

The derivative of the constant $\log 2$ is $0$, and the derivative of $-\sin (u)$ is $-\cos (u)$. Thus:

Differentiating the inner part, $\frac{d}{dx}(\log 2-\sin x)$, gives us $-\cos x$. Combining these, the two negative signs cancel out:

We are at the finish line. We need to evaluate $g^{\prime }(0)$ by substituting $x=0$ into our derivative:

Since $\sin 0=0$ and $\cos 0=1$, the expression simplifies significantly. The final answer is:

You have successfully navigated the modulus, the composite function, and the Chain Rule. This is the essence of JEE Advanced mathematics—breaking down complex structures into simple, elegant truths.