Q.Find $\frac{dy}{dx}$ in the following: $(x+3)^2 \cdot (x+4)^3 \cdot (x+5)^4$

The key idea is Implicit Differentiation — but here we have an explicit product of three factors, so we use logarithmic differentiation (a direct application of implicit differentiation) to handle the product of powers efficiently.

Take the natural log of both sides:

Differentiate both sides with respect to $x$:

Multiply through by $y$ and substitute back the original expression: …

Why logarithmic differentiation?

When you have a function that is a product of several factors, each raised to a power, the standard approach (product rule repeated) becomes messy fast. Here, we have three factors: $(x+3)^2$, $(x+4)^3$, and $(x+5)^4$. Applying the product rule directly would require differentiating a product of three functions, each of which itself requires the chain rule. It's doable but tedious.

Logarithmic differentiation turns multiplication into addition. By taking the natural logarithm of both sides, the powers become coefficients, and the product becomes a sum — much easier to differentiate. Then we just solve for $\frac{dy}{dx}$.

Let's work through it.

1. Set up the equation. Let

2. Take the natural logarithm of both sides. This is valid because $y > 0$ for all $x > -3$ (the domain where all factors are positive; for other $x$, we can take absolute values, but the derivative formula will still hold).

3. Use logarithm properties to simplify. The log of a product is the sum of logs, and the log of a power brings the exponent down:

Now we have a sum of simple logarithmic terms — much easier to differentiate.

4. Differentiate both sides with respect to $x$. On the left, by the chain rule:

On the right, differentiate term by term:

So we have: …

Method: Logarithmic Differentiation for a Product of Powers with Constant Exponents

Use this method when $y$ is a product of several binomial (or other) factors, each raised to a fixed, constant power — here $(x+3)^2(x+4)^3(x+5)^4$. Multiplying out or applying the product rule three times would be extremely tedious.

Steps

Step 1: Take the natural log of both sides and expand using log laws.

Each constant exponent drops down and becomes a multiplying coefficient in front of its own log term.

Step 2: Differentiate term by term.

Since the derivative of each linear factor $(x+a)$ is simply $1$, each term differentiates to a plain reciprocal times its coefficient:

$\frac{1}{y}\frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}$ …

Common Mistakes

Mistake 1: Confusing this with a case needing an extra Chain Rule factor.

Why it's wrong: because $\frac{d}{dx}(x+a) = 1$ for these simple linear factors, no extra multiplier appears beyond the constant exponent itself; some students, out of habit from harder problems, multiply by an extra (unneeded) derivative term. Correct approach: always differentiate the inner expression first — here it genuinely equals $1$, so the coefficient in each term is exactly the original exponent.

Mistake 2: Forgetting to bring the exponent down as a coefficient. …