y= [3] 1+3 x [4] 1+4 x [5] 1+5 x [7] 1+7 x [8] 1+8 x . Then d y ~d x at x=0 is

Correct answer: C. 1 aligned +(1)/(5) (1+5 x) & -(1)/(7) (1+7 x) & -(1)/(8) (1+8 x) aligned Differentiating both sides w.r.t. x , we get array r (1)/(y) · d y ~d x =(1)/(3) · (1)/(1+3 x) ·₃⁺(1)/(4) · (1)/(1+4 x) ·₄⁺(1)/(5) · (1)/(1+5 x) ·5 -(1)/(7) · (1)/(1+7 x) ·₇⁻(1)/(8) · (1)/(1+8 x) ·8 array aligned & (1)/(y) · d y ~d x =(1)/(1+3 x)+(1)/(1+4 x)+(1)/(1+5 x)-(1)/(1+7 x)-(1)/(1+8 x) & (1)/(1) ·( d y ~d x )_ x=0 =(1)/(1+0)+(1)/(1+0)+(1)/(1+0)-(1)/(1+0)-(1)/(1+0) & [ At x=0, y=1] aligned ( d y ~d x )_ x=0 =1

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Differentiation

JEE Main · Mathematics · Differentiation

y= [3] 1+3 x [4] 1+4 x [5] 1+5 x [7] 1+7 x [8] 1+8 x . Then d y ~d...

y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}

. Then

\frac{\mathrm{d} y}{\mathrm{~d} x}

x=0

Options

A3
B $-1$
C1
D2

Correct answer

C. 1

Step-by-step solution

\begin{aligned}+\frac{1}{5} \log (1+5 x) & -\frac{1}{7} \log (1+7 x) \\ & -\frac{1}{8} \log (1+8 x)\end{aligned}

Differentiating both sides w.r.t.

x

, we get

\begin{array}{r}\frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{3} \cdot \frac{1}{1+3 x} \cdot₃⁺\frac{1}{4} \cdot \frac{1}{1+4 x} \cdot₄⁺\frac{1}{5} \cdot \frac{1}{1+5 x} \cdot5 \\ -\frac{1}{7} \cdot \frac{1}{1+7 x} \cdot₇⁻\frac{1}{8} \cdot \frac{1}{1+8 x} \cdot8 \end{array}

\begin{aligned} & \therefore \quad \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+3 x}+\frac{1}{1+4 x}+\frac{1}{1+5 x}-\frac{1}{1+7 x}-\frac{1}{1+8 x} \\ & \therefore \quad \frac{1}{1} \cdot\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\frac{1}{1+0}+\frac{1}{1+0}+\frac{1}{1+0}-\frac{1}{1+0}-\frac{1}{1+0} \\ & \ldots[\text { At } x=0, y=1]\end{aligned}

\Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=1

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