Animated Solution for Mathematics - Matrices and Determinants: If $f(x)=\left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3+2{\cos }^{4}x& 2{\sin }^{4}x& {\sin }^{2}2x\\ 2{\cos }^{4}x& 3+2{\sin }^{4}x& {\sin }^{2}2x\end{array}\right|$ then $\frac{1}{5}{f}^{\prime }(0)$ is equal to

Welcome, future IITian! Today we are going to conquer a problem that looks like a nightmare but is actually a beautiful dance of symmetry.

When you first look at the determinant

your instinct might be to panic. It is filled with trigonometric powers and looks like it would take an hour to differentiate. But remember, in the JEE Advanced arena, complexity is often just a mask for elegance.

Before we touch any calculus, we must simplify. Look at the rows; the $2{\cos }^{4}x$ and $2{\sin }^{4}x$ terms are begging to be cancelled.

We use the power of row operations: $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$.

When we subtract the first row from the second, the $2{\cos }^{4}x$ terms cancel, leaving $3$, the $2{\sin }^{4}x$ terms cancel to $0$, and the ${\sin }^{2}2x$ terms leave us with $-3$. Repeating this for the third row, we generate another set of simple constants.

Our determinant transforms into:

Suddenly, the beast is tamed.

Now, we expand along the second row. It is the most efficient path because of the zero.

The expansion gives us:

Evaluating these minors carefully, we get:

Distributing the constants, we arrive at:

Grouping the terms, we see $18({\sin }^{4}x+{\cos }^{4}x)+9{\sin }^{2}2x+27$.

Here is the magic trick. We recall the identity ${\sin }^{4}x+{\cos }^{4}x=1-\frac{1}{2}{\sin }^{2}2x$.

Substituting this back, we get:

Expanding this, the ${\sin }^{2}2x$ terms cancel out entirely! We are left with $f(x)=18+27=45$.

The entire function is just a constant. Therefore, $f^{\prime }(x)=0$, and the final result is:

You see? The complexity was just a test of your patience and your ability to see the underlying structure. Keep practicing this, and you will start seeing these patterns everywhere!