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

$y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}$. Then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ is
  1. 3
  2. $-1$
  3. 1
  4. 2

Solution

$y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}$ $\therefore \quad \log y=\frac{1}{3} \log (1+3 x)+\frac{1}{4} \log (1+4 x)$ $\begin{aligned}+\frac{1}{5} \log (1+5 x) & -\frac{1}{7} \log (1+7 x) \\ & -\frac{1}{8} \log (1+8 x)\end{aligned}$ Differentiating both sides w.r.t. $x$, we get $\begin{array}{r} \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{3} \cdot \frac{1}{1+3 x} \cdot 3+\frac{1}{4} \cdot \frac{1}{1+4 x} \cdot 4+\frac{1}{5} \cdot \frac{1}{1+5 x} \cdot 5 \\ -\frac{1}{7} \cdot \frac{1}{1+7 x} \cdot 7-\frac{1}{8} \cdot \frac{1}{1+8 x} \cdot 8 \end{array}$ $\begin{aligned} & \therefore \quad \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+3 x}+\frac{1}{1+4 x}+\frac{1}{1+5 x}-\frac{1}{1+7 x}-\frac{1}{1+8 x} \\ & \therefore \quad \frac{1}{1} \cdot\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\frac{1}{1+0}+\frac{1}{1+0}+\frac{1}{1+0}-\frac{1}{1+0}-\frac{1}{1+0} \\ & \ldots[\text { At } x=0, y=1]\end{aligned}$ $\Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=1$

Asked in: MHT CET 2023 (13 May Shift 2)

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