Q.Find $\frac{dy}{dx}$ in the following: $\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Concept: Implicit Differentiation — but here we use logarithmic differentiation (take logs first, then differentiate implicitly) because the function is a complicated product/quotient under a square root.

Let

Step 1: Take natural log on both sides and simplify using log laws.

Step 2: Differentiate both sides with respect to $x$ (implicitly on the left).

Step 3: Multiply both sides by $y$ and substitute back the original $y$.

The problem asks for $\frac{dy}{dx}$ when $y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$. This is a function built from products and quotients of linear factors, all under a square root. The direct approach — applying quotient rule, product rule, and chain rule — would be a nightmare of algebra. There’s a cleaner way.

The core idea: When a function is a product or quotient of powers, take the natural logarithm first. The logarithm turns multiplication into addition and division into subtraction. Then differentiate implicitly. This is called logarithmic differentiation, and it’s the standard tool for problems like this.

Let’s do it step by step.

1. Set up the equation. Let $y = \left[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right]^{1/2}$. Take the natural log of both sides:

2. Expand the log using log laws. The log of a quotient is the difference of logs, and the log of a product is the sum of logs:

This is the key simplification. Instead of a complicated fraction, we now have a sum of simple terms.

3. Differentiate both sides with respect to $x$. On the left, by implicit differentiation: $\frac{d}{dx}[\log y] = \frac{1}{y} \frac{dy}{dx}$. On the right, differentiate term-by-term. Remember $\frac{d}{dx}[\log(x-a)] = \frac{1}{x-a}$:

4. Solve for $\frac{dy}{dx}$. Multiply both sides by $y$:

That’s the derivative in compact form. If you want it fully in terms of $x$, substitute back the original expression for $y$:

Method: Logarithmic Differentiation for a Quotient of Products (Including Roots)

Use this method whenever $y$ involves a product or quotient of several linear (or simple) factors, especially when a root is also present — direct use of the quotient rule and product rule together would be extremely messy.

Steps

Step 1: Rewrite any root as a fractional power, then take logs of both sides.

The exponent $\frac12$ comes outside as a multiplier; every factor in the numerator contributes a $+\log(\cdot)$ term, and every factor in the denominator contributes a $-\log(\cdot)$ term.

Step 2: Differentiate both sides.

Each term is of the form $\log(x-a)$, whose derivative is simply $\frac{1}{x-a}$ (the inner derivative is $1$):

Step 3: Multiply by $y$ and substitute the original expression back in.

The general pattern: for $y=\left[\dfrac{\prod(\text{numerator factors})}{\prod(\text{denominator factors})}\right]^{n}$, the derivative is always $n\cdot y \times$ (sum of reciprocals of numerator factors minus sum of reciprocals of denominator factors).

Common Mistakes

Mistake 1: Treating denominator factors the same as numerator factors (sign error).

Why it's wrong: because $\log\frac{A}{B} = \log A - \log B$, the three denominator factors $(x-3)$, $(x-4)$, $(x-5)$ must enter with a minus sign, not a plus. Adding all five terms with the same sign is a very common slip. Correct approach: split into numerator-log-terms (positive) and denominator-log-terms (negative) before differentiating.

Mistake 2: Forgetting the $\frac12$ factor from the square root.

Why it's wrong: a square root is a power of $\frac12$; skipping this when expanding the log means every term in the final derivative is twice as large as it should be. Correct approach: rewrite the root as a fractional exponent before taking logs, so the $\frac12$ is visible and gets multiplied through every term.

Mistake 3: Not restricting the domain / ignoring where the expression is undefined.

Why it's wrong: the original function is undefined at $x=3,4,5$ (denominator zero) and requires the expression under the root to be non-negative, so the derivative formula is only valid on the domain where $y$ itself is defined. Correct approach: state the natural domain restrictions alongside the derivative rather than treating the formula as valid for all $x$.