Animated Solution for Mathematics - Limits, Continuity and Differentiability: Suppose $f(x)=\frac{(2^x+2^{-x})\tan x\sqrt{{\tan }^{-1}(x^2-x+1)}}{(7x^2+3x+1{)}^3}$. Then the value of $f^{\prime }(0)$ is equal to

The function provided is:

If you attempt to differentiate this using the standard quotient rule, you are walking into a trap. The examiners are testing your intuition to recognize when to apply the First Principle of Derivatives.

First, we must evaluate $f(0)$. When we substitute $x=0$ into the expression:

The term $\tan (0)$ is $0$. Since this is a product, the entire numerator collapses to $0$.

The denominator becomes $(7(0{)}^2+3(0)+1{)}^3$, which simplifies to $1^3=1$. Thus, we find that $f(0)=0$.

We now invoke the First Principle of Derivatives:

Since $f(0)=0$, this expression simplifies beautifully to:

Look closely at the expression for $f(h)$. We can isolate the term $\frac{\tan h}{h}$, which is a standard limit known to equal $1$:

By isolating this, we have removed the indeterminate form. The remaining terms can now be evaluated directly as $h\to 0$.

Evaluating the remaining components as $h\to 0$:

The exponential term becomes $2^0+2^{-0}=1+1=2$.

The square root term becomes $\sqrt{{\tan }^{-1}(0^2-0+1)}=\sqrt{{\tan }^{-1}(1)}=\sqrt{\frac{\pi }{4}}=\frac{\sqrt{\pi }}{2}$.

The denominator simplifies to $1^3=1$. Multiplying these pieces together:

The final result is $\sqrt{\pi }$.

The lesson here is clear: in JEE Advanced, the most complex-looking problems often yield to the most elegant solutions if you look past the surface. Never rush into brute force; analyze the structure and find the path of least resistance.