A) \[\sin \left[ \log \left( \frac{2x+3}{3-2x} \right) \right]\]
B) \[\frac{12}{{{\left( 3-2x \right)}^{2}}}\]
C) \[\frac{12}{{{\left( 3-2x \right)}^{2}}}\sin \left[ \log \left( \frac{2x+3}{3-2x} \right) \right]\]
D) \[\frac{12}{{{\left( 3-2x \right)}^{2}}}\cos \left[ \log \left( \frac{2x+3}{3-2x} \right) \right]\]
Correct Answer: C
Solution :
Let\[f'(x)=\sin [\log x]\]and\[y=f\left( \frac{2x+3}{3-2x} \right)\] Now,\[\frac{dy}{dx}=f'\left( \frac{2x+3}{3-2x} \right).\frac{d}{dx}\left( \frac{2x+3}{3-2x} \right)\] \[=\sin \left[ \log \left( \frac{2x+3}{3-2x} \right) \right]\frac{\left[ \left( 6-4x \right)-\left( -4-6 \right) \right]}{{{\left( 3-2x \right)}^{2}}}\] \[=\frac{12}{{{\left( 3-2x \right)}^{2}}}.\sin \left[ \log \left( \frac{2x+3}{3-2x} \right) \right]\]
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