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

Question

Dr. Abubacker · Accepted Answer

(c)
$\frac{dy}{dx} = f^{'}\left( \frac{2x + 3}{3 - 2x} ight)\frac{d}{dx}\ \left( \frac{2x + 3}{3 - 2x} ight)$
$= {\ sin}{\log\left( \frac{2x + 3}{3 - 2x} ight)\left( \frac{(3 - 2x)(2) - (2x + 3)( - 2)}{(3 - 2x)^{2}} ight)}$
$= \frac{12}{(3 - 2x)^{2}}\sin\left\{ \log\left( \frac{2x + 3}{3 - 2x} ight) ight\}$
$\ 	herefore\ \ \left( \frac{dy}{dx} ight)_{x = 1} = \frac{12}{(3 - 2)^{2}}.\sin{\log{5 = 12\sin{\log{5\ }}}}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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