Question

If $ f'(x)=\sin (\log x) $ and $ y=f\left( \frac{2x+3}{3-2x} \right), $ then $ \frac{dy}{dx} $ at $ x=1 $ is equal to:

• A. 6 sin log (5)
• B. 5 sin log (6)
• C. 12 sin log (5)
• D. 5 sin log (12)
• E. 12 sin log (6)

Answer 1

Answer: (C)

$ \frac{dy}{dx}=f\left( \frac{2x+3}{3-2x} \right)\frac{d}{dx}\left( \frac{2x+3}{3-2x} \right) $
$ =\sin \log \left( \frac{2x+3}{3-2x} \right)\left( \frac{(3-2x)(2)-(2x+3)(-2)}{{{(3-2x)}^{2}}} \right) $ $ =\frac{12}{{{(3-2x)}^{2}}}\sin \left\{ \log \left( \frac{2x+3}{3-2x} \right) \right\} $
$ \therefore $ $ {{\left( \frac{dy}{dx} \right)}_{x=1}}=\frac{12}{{{(3-2)}^{2}}}.\sin \,\log 5 $
= 12 sin log 5