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JEE Main 201212 MayOnline
Exam
JEE Main
Year
2012
Date
12 May
Mode
Online
Subject
Mathematics
Chapter
Differentiation

JEE Main 2012 · 12 May · Online · Mathematics · Differentiation

If f^ (x)= ( x) and y=f((2 x+3)/(3-2 x)) , then (d y)/(d x) equals

If f′(x)=sin⁡(log⁡x)f^{\prime}(x)=\sin (\log x) and y=f(2x+33−2x)y=f\left(\frac{2 x+3}{3-2 x}\right), then dydx\frac{d y}{d x} equals

Options

  1. Asin⁡[log⁡(2x+33−2x)]\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]
  2. B\frac{12}{\left(3-2 x^2\right)}
  3. C12(3−2x2)sin⁡[log⁡(2x+33−2x)]\frac{12}{\left(3-2 x^2\right)} \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]
  4. D12(3−2x2cos⁡[log⁡(2x+33−2x)]\frac{12}{\left(3-2 x^2\right.} \cos \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]

Correct answer

C. 12(3−2x2)sin⁡[log⁡(2x+33−2x)]\frac{12}{\left(3-2 x^2\right)} \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]

Step-by-step solution

Let f′(x)=sin⁡[log⁡x]andf^{\prime}(x)=\sin [\log x] andy=f \left(\frac{2 x+3}{3-2 x}\right) Now, \frac{d y}{d x}=f′(2x+33−2x)⋅ddx(2x+33−2x)f^{\prime} \left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)=sin⁡[log⁡(2x+33−2x)][(6−4x−)−4x−6](3−2x2)=12(3−2x2⋅sin⁡[log⁡(2x+33−2x)]\begin{aligned}& =\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right] \frac{[(6-4 x-)-4 x-6]}{\left(3-2 x^2\right)}\\ & =\frac{12}{\left(3-2 x^2\right.} \cdot \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right] \end{aligned}$$
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