MHT CET · Maths · Differentiation
If \(\mathrm{f}^{\prime}(x)=\sin (\log x)\) and \(y=\mathrm{f}\left(\frac{2 x+3}{3-2 x}\right)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=1\) is
- A \(6 \sin (\log 5)\)
- B \(5 \sin (\log 6)\)
- C \(12 \sin (\log 5)\)
- D \(5 \sin (\log 12)\)
Answer & Solution
Correct Answer
(C) \(12 \sin (\log 5)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\mathrm{f}\left(\frac{2 x+3}{3-2 x}\right) \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{f}^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{2 x+3}{3-2 x}\right) \\ & =\mathrm{f}^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{(3-2 x) \cdot 2-(2 x+3)(-2)}{(3-2 x)^2} \\ & =\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right] \cdot \frac{12}{(3-2 x)^2} \\ & \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=1}=\sin (\log 5) \cdot 12=12 \sin (\log 5) \\ & \end{aligned}\)
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