Animated Solution for Mathematics - Differentiation: If $y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{x-c}+1$, prove that $\frac{y^{\prime }}{y}=\frac{1}{x}\left[\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\right]$.

Imagine you are standing before a massive, intimidating algebraic structure. It looks like a fortress of fractions, products, and variables.

Your instinct might be to charge at it with the quotient rule, but in the world of JEE Advanced, brute force is rarely the path to victory. Today, we are going to dismantle this 'monster' expression not by force, but by strategy.

The Illusion of Complexity

Our starting expression is:

If you try to differentiate this directly, you will be trapped in a labyrinth of product and quotient rules. Instead, let's look at the tail end of the expression.

Notice the last two terms: $\frac{c}{x-c}+1$. When we combine these by taking the lowest common multiple, we get:

Look at that! The $c$ and $-c$ vanish, leaving us with a simple $\frac{x}{x-c}$. This is the first crack in the fortress walls.

The Telescoping Pattern

Now, we take our result, $\frac{x}{x-c}$, and add it to the next term: $\frac{bx}{(x-b)(x-c)}$. Again, we find a common denominator, which is $(x-b)(x-c)$:

Again, the $bx$ and $-bx$ cancel out perfectly. Do you see the pattern? The expression is collapsing.

If we continue this process with the first term, $\frac{a{x}^2}{(x-a)(x-b)(x-c)}$, we eventually arrive at a beautifully compact form:

The Logarithmic Shortcut

Now that we have a clean expression for $y$, we need to find $\frac{y^{\prime }}{y}$. Instead of using the quotient rule, we use the magic wand of calculus: logarithmic differentiation.

By taking the natural logarithm of both sides, we turn the multiplication and division into simple addition and subtraction:

Using the properties of logarithms, this expands to:

Now, differentiating with respect to $x$ becomes a breeze:

The Final Symmetry

We are almost at the finish line. The expression we need to prove is $\frac{y^{\prime }}{y}=\frac{1}{x}\left[\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\right]$.

Our current expression has a $\frac{3}{x}$, but we need to distribute it. We split $\frac{3}{x}$ into $\frac{1}{x}+\frac{1}{x}+\frac{1}{x}$ and group them:

Focus on the first pair: $\frac{1}{x}-\frac{1}{x-a}=\frac{(x-a)-x}{x(x-a)}=\frac{-a}{x(x-a)}$. By flipping the denominator to $a-x$, we get $\frac{a}{x(a-x)}$.

Applying this logic to all three pairs, we get:

Finally, factoring out $\frac{1}{x}$ gives us exactly what we set out to prove:

And there you have it. We didn't fight the monster; we simplified it until it revealed its true, elegant form. Keep this mindset—look for the pattern, simplify, and then execute.