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Question: If f ў(x) = sin (log x) and y = ѓ\(\left( \frac{2x + 3}{3 - 2x} \right)\), then \(\frac{dy}{dx}\) is...

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

A

9cos⁡(log⁡x)x(3−2x)2\frac{9\cos(\log x)}{x(3 - 2x)^{2}}

B

9cos⁡(log⁡2x+33−2x)x(3−2x)2\frac{9\cos\left( \log\frac{2x + 3}{3 - 2x} \right)}{x(3 - 2x)^{2}}

C

9sin⁡(log⁡(2x+33−2x))(3−2x)2\frac{9\sin\left( \log\left( \frac{2x + 3}{3 - 2x} \right) \right)}{(3 - 2x)^{2}}

D

None of these

Answer

None of these

Explanation

Solution

dydx\frac{dy}{dx} = ƒ¢(z) dzdx\frac{dz}{dx} …(1)

Where z = 2x+33−2x\frac{2x + 3}{3 - 2x},

dzdx\frac{dz}{dx} = (3−2x).2−(2x+3).(–2)(3–2x)2\frac{(3 - 2x).2 - (2x + 3).(–2)}{(3–2x)^{2}}

= 12(3−2x)2\frac{12}{(3 - 2x)^{2}}

\ From (1), dydx\frac{dy}{dx} = sin (log z) dzdx\frac{dz}{dx}

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