Q.Find the derivative of the function given by $f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)$ and hence find $f'(1)$.

Concept: Chain Rule & Logarithmic Differentiation – when a product has many factors, taking logs simplifies the derivative.

Step 1: Take the natural logarithm of both sides:

Step 2: Differentiate both sides with respect to $x$:

Step 3: Multiply through by $f(x)$:

$f'(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right)$ …

Solution

1. Telescope the product.

Multiply $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$ by $(1-x)$ and use difference of squares repeatedly:

Hence $(1-x)f(x)=1-x^{16}$, i.e. for $x\neq1$

(As a degree‑15 polynomial, this identity extends to $x=1$ by continuity.)

2. Differentiate.

3. Evaluate at $x=1$.

$f'(1)=1+2+3+\cdots+15=\frac{15\cdot 16}{2}=120.$ …

Method: Logarithmic Differentiation to Handle a Product of Several Factors, Then Evaluate at a Point

When a function is a product of more than two factors, repeated product rule gets long fast. Logarithmic differentiation turns the product into a sum before you differentiate, which is far less error-prone — especially when you only need the derivative's value at one specific point.

Steps

Step 1: Take the log of the whole product

For $f(x) = f_1(x)f_2(x)\cdots f_n(x)$:

The product has become a sum — much easier to differentiate term by term.

Step 2: Differentiate both sides

Each term on the right is a simple ratio — compute them independently.

Step 3: Multiply through by $f(x)$

$f'(x) = f(x)\left(\frac{f_1'(x)}{f_1(x)} + \cdots + \frac{f_n'(x)}{f_n(x)}\right)$ …

Common Mistakes

Mistake 1: Forgetting to multiply back by $f(x)$ after differentiating $\log f(x)$

Why it's wrong: logarithmic differentiation gives you $\dfrac{f'(x)}{f(x)}$ directly, not $f'(x)$ itself — a student who computes the bracket of fractions correctly but then reports that value alone (without multiplying by $f(x)$) is off by a factor of $f(1)=16$ in this problem. Correct approach: always write the final line as $f'(x) = f(x)\times(\text{the bracket})$ before substituting the point.

Mistake 2: Arithmetic slip evaluating each fraction at $x=1$ …