If f ′ ( x ) = sin ( lo g x ) and y = f ( 3 − 2 x 2 x + 3 ) , then d x d y equals

Let f ′ ( x ) = sin [ lo g x ] and y = f ( 3 − 2 x 2 x + 3 ) Now, d x d y = f ′ ( 3 − 2 x 2 x + 3 ) ⋅ d x d ( 3 − 2 x 2 x + 3 ) = sin [ lo g ( 3 − 2 x 2 x + 3 ) ] ( 3 − 2 x 2 ) [( 6 − 4 x − ) − 4 x − 6 ] = ( 3 − 2 x 2 12 ⋅ sin [ lo g ( 3 − 2 x 2 x + 3 ) ]

Mathematics - Differentiation Question with Solution | TestHub

MathematicsDifferentiationDifferentiation of composite functionsHard2 minPYQ_2012

MathematicsHardsingle choice

If $f^{'} (x) = sin (lo g x)$ and $y = f (\frac{2 x + 3}{3 - 2 x})$ , then $\frac{d y}{d x}$ equals

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Let $f^{'} (x) = sin [lo g x]$ and $y = f (\frac{2 x + 3}{3 - 2 x})$ Now, $\frac{d y}{d x} = f^{'} (\frac{2 x + 3}{3 - 2 x}) \cdot \frac{d}{d x} (\frac{2 x + 3}{3 - 2 x})$ $= sin [lo g (\frac{2 x + 3}{3 - 2 x})] \frac{[( 6 - 4 x - ) - 4 x - 6 ]}{( 3 - 2 x ^{2} )} = \frac{12}{( 3 - 2 x ^{2}} \cdot sin [lo g (\frac{2 x + 3}{3 - 2 x})]$

Stream:JEESubject:MathematicsTopic:DifferentiationSubtopic:Differentiation of composite functions

⏱ 2mℹ️ Source: PYQ_2012

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