If y=[(x+1)(2 x+1)(3 x+1) . .(n x+1)]² then d y/d x at x=0 is

Correct Answer is : n(n+1){n}({n}+1)n(n+1)

Correct Answer is : n(n+1){n}({n}+1)n(n+1)

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MHTCET 15 May 2024 Shift 1 Paper

Question : 101 of 150

Marks: +1, -0

y=[(x+1)(2 x+1)(3 x+1) \ldots \ldots . .(n x+1)]^{2}

\frac{d y}{d x}

x=0

Solution:

Concept:

Use logarithmic differentiation to differentiate a product raised to a power.

Explanation:

Given

y=[(x+1)(2x+1)(3x+1)\ldots(nx+1)]^{2}

, take natural log on both sides.

\log y = 2[\log(x+1)+\log(2x+1)+\log(3x+1)+\ldots+\log(nx+1)]

Differentiate both sides with respect to

x

\frac{1}{y}\frac{dy}{dx} = 2\left(\frac{1}{x+1}+\frac{2}{2x+1}+\frac{3}{3x+1}+\ldots+\frac{n}{nx+1}\right)

So,

\frac{dy}{dx} = 2y\left(\frac{1}{x+1}+\frac{2}{2x+1}+\ldots+\frac{n}{nx+1}\right)

x=0

, each factor inside the product is

1

, hence

y=1

Substitute

x=0

\left(\frac{dy}{dx}\right)_{x=0} = 2(1)(1+2+3+\ldots+n)

= 2 \times \frac{n(n+1)}{2} = n(n+1)

Answer:

n(n+1)

, which is option B.

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