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Mathematics Question on Continuity and differentiability

If f(x) = sin (log x) and y=f(2x+33−2x)y = f\left(\frac{2x+3}{3-2x}\right), then dydx\frac{dy}{dx} equals

A

sin[log(2x+33−2x)]sin\left[log\left(\frac{2x+3}{3-2x}\right)\right]

B

12(3−2x)2\frac{12}{\left(3-2x\right)^2}

C

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

D

12(3−2x)2cos[log(2x+33−2x)]\frac{12}{\left(3-2x\right)^{2}}cos\left[log\left(\frac{2x+3}{3-2x}\right)\right]

Answer

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

Explanation

Solution

Let f′(x)=sin[logx]f'\left(x\right) = sin \left[log x\right] and y=f(2x+33−2x)y = f\left(\frac{2x+3}{3-2x}\right) Now, dydx=f′(2x+33−2x).ddx(2x+33−2x)\frac{dy}{dx} = f' \left(\frac{2x+3}{3-2x}\right) . \frac{d}{dx}\left(\frac{2x+3}{3-2x}\right) =sin[log(2x+33−2x)][(6−4x−)−4x(−6)](3−2x2)= sin\left[log\left(\frac{2x+3}{3-2x}\right)\right] \frac{\left[\left(6-4x-\right)-4x\left(-6\right)\right]}{\left(3-2x^{2}\right)} =12(3−2x2)sin[log(2x+33−2x)]= \frac{12}{\left(3-2x^{2}\right)}sin\left[log\left(\frac{2x+3}{3-2x}\right)\right]