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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Differentiation

BITSAT 2023 · Paper 2 · Mathematics · Differentiation

If y^x=e^ y-x , then (d y)/(d x) is equal to

If yx=ey−xy^x=e^{y-x}, then dydx\frac{d y}{d x} is equal to

Options

  1. A1+log⁡yylog⁡y\frac{1+\log y}{y \log y}
  2. B(1+log⁡y)2ylog⁡y\frac{(1+\log y)^2}{y \log y}
  3. C1+log⁡y(log⁡y)2\frac{1+\log y}{(\log y)^2}
  4. D(1+log⁡y)2log⁡y\frac{(1+\log y)^2}{\log y}

Correct answer

D. (1+log⁡y)2log⁡y\frac{(1+\log y)^2}{\log y}

Step-by-step solution

Here, yx=ey−x\mathrm{y}^{\mathrm{x}}=\mathrm{e}^{\mathrm{y}-\mathrm{x}}Taking log on both sides, we getlog⁡yx=log⁡ey−x(∵log⁡ab=blog⁡a and  log⁡e=1)\begin{aligned} \\ & \log y^x=\log e^{y-x} \\ \\ & \quad\left(\because \log a^b=b \log a \text{\text{ and } } \log e=1\right) \\ \end{aligned}\Rightarrow x \log y=(y-x) \log e \Rightarrow x \log y=y-x \ldots(i)Ondifferentiatingw.r.t.On differentiating w.r.t.x,weget, we get\frac{d}{d x}(x \log y)=\frac{d}{d x}(y-x)(usingproductrule)(using product rule)⇒x(1y)dydx+log⁡y(1)=dydx−1⇒dydx(xy−1)=−1−log⁡y⇒dydx[y(1+log⁡y)y−1]=−(1+log⁡y)[∵ from  eq. (i)⋅x=y(1+log⁡y)]⇒dydx[1−1−log⁡y1+log⁡y]=−(1+log⁡y)\begin{aligned} \\ & \Rightarrow x\left(\frac{1}{y}\right) \frac{d y}{d x}+\log y(1)=\frac{d y}{d x}-1 \\ \\ & \Rightarrow \frac{d y}{d x}\left(\frac{x}{y}-1\right)=-1-\log y \\ \\ & \Rightarrow \frac{d y}{d x}\left[\frac{y}{(1+\log y) y}-1\right]=-(1+\log y) \\ \\ & \qquad \quad\left[\because \text{\text{ from } eq. }(i) \cdot x=\frac{y}{(1+\log y)}\right] \\ \\ & \Rightarrow \frac{d y}{d x}\left[\frac{1-1-\log y}{1+\log y}\right]=-(1+\log y) \\ \end{aligned}⇒dydx=−(1+log⁡y)2−log⁡y⇒dydx=(1+log⁡y)2log⁡y\Rightarrow \frac{d y}{d x}=-\frac{(1+\log y)^2}{-\log y} \Rightarrow \frac{d y}{d x}=\frac{(1+\log y)^2}{\log y}
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