Question

If $y = \left[(x+1)(2x+1)(3x+1)\cdots(nx+1)\right]^n$, then $\frac{dy}{dx}$ at $x=0$ is

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n^{2}(n+1)}{2}$
• C. $\frac{n(n+1)}{4}$
• D. $\frac{n^{2}(n-1)}{2}$

Hint:

Calculus Tip: When differentiating the product of many terms, logarithmic differentiation is the fastest and most reliable method to turn massive product rules into simple sum rules.

Answer 1

The Correct Option is B

Concept: Calculus - Logarithmic Differentiation.

Step 1: Apply natural logarithm to both sides of the equation. Given $y = [(x+1)(2x+1)(3x+1)...(nx+1)]^n$. Take the natural log ($\ln$ or $\log$) on both sides: $\log y = \log \left( [(x+1)(2x+1)...(nx+1)]^n \right)$. Using the power rule of logarithms ($\log(A^B) = B\log A$): $\log y = n \log[(x+1)(2x+1)...(nx+1)]$.

Step 2: Expand the logarithmic expression. Use the product rule of logarithms ($\log(AB) = \log A + \log B$) to expand the terms inside the bracket: $\log y = n [\log(x+1) + \log(2x+1) + \log(3x+1) + ... + \log(nx+1)]$.

Step 3: Differentiate implicitly with respect to $x$. Differentiate both sides with respect to $x$. The derivative of $\log y$ is $\frac{1}{y} \frac{dy}{dx}$. Apply the chain rule to the right side: $\frac{1}{y} \frac{dy}{dx} = n \left[ \frac{1}{x+1}(1) + \frac{1}{2x+1}(2) + \frac{1}{3x+1}(3) + ... + \frac{1}{nx+1}(n) \right]$.

Step 4: Evaluate $y$ at $x=0$. Before substituting $x=0$ into our derivative equation, we need to know the value of $y$ at $x=0$. Substitute $x=0$ into the original equation: $y(0) = [(0+1)(0+1)(0+1)...(0+1)]^n = [1 \cdot 1 \cdot 1 ... 1]^n = 1^n = 1$.

Step 5: Substitute $x=0$ and solve for the derivative. Substitute $x=0$ and $y=1$ into the differentiated equation from
Step 3: $\frac{1}{1} \left( \frac{dy}{dx} \right)_{x=0} = n \left[ \frac{1}{0+1} + \frac{2}{0+1} + \frac{3}{0+1} + ... + \frac{n}{0+1} \right]$. $\left( \frac{dy}{dx} \right)_{x=0} = n [1 + 2 + 3 + ... + n]$. The sum of the first $n$ natural numbers is given by the formula $\frac{n(n+1)}{2}$. Therefore, $\left( \frac{dy}{dx} \right)_{x=0} = n \left[ \frac{n(n+1)}{2} \right] = \frac{n^2(n+1)}{2}$. $$ \therefore \text{The value of } \frac{dy}{dx} \text{ at } x=0 \text{ is } \frac{n^{2}(n+1)}{2}. $$