First slide

Differentiability

Question

$If f^{1} (x) = \sin (\log x) and y = f (\frac{2 x + 3}{3 - 2 x}), then \frac{d y}{d x} at x = 1 is equal to (Base of all \log = e)$

Moderate

A

$6 sinlog (5)$
B

$5 sinlog (6)$
C

$12 sinlog (5)$
D

$5 sinlog (12)$

Solution

$\begin{array}{l} Given that f^{'} (x) = \sin (\log x) and Given that y = f (\frac{2 x + 3}{3 - 2 x}) \\ \Rightarrow \frac{d y}{d x} = f^{'} (\frac{2 x + 3}{3 - 2 x}) \frac{d}{d x} (\frac{2 x + 3}{3 - 2 x}) \\ = \sin (\log (\frac{2 x + 3}{3 - 2 x})) \cdot [\frac{(3 - 2 x) (2) - (2 x + 3) (- 2)}{(3 - 2 x)^{2}}] \\ = \frac{12}{(3 - 2 x)^{2}} \sin [\log (\frac{2 x + 3}{3 - 2 x})] \\ ∴ {(\frac{d y}{d x})}_{(x = 1)} = \frac{12}{(3 - 2)^{2}} sinlog (5) \\ = 12 \sin (\log 5) \end{array}$

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