7.3.29-eg - e-4 t (-sin(2 t )) (2) thus dy dt =-12 e-4 t...

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Example Find y 0 if y = ln(3 e - 4 t cos(2 t )). First note that y = f ( g ( t )), where f ( t ) = ln( t ) and g ( t ) = 3 e - 4 t cos(2 t ), so by chain rule dy dt = f 0 ( g ( t )) g 0 ( t ) . For this problem, f 0 ( t ) = 1 t and g 0 ( t ) = d dt (3 e - 4 t ) · cos(2 t ) + 3 e - 4 t · d dt (cos(2 t )) = 3 e - 4 t · ( - 4) cos(2 t ) + 3
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Unformatted text preview: e-4 t (-sin(2 t )) (2) thus dy dt =-12 e-4 t cos(2 t )-6 e-4 t sin(2 t ) 3 e-4 t cos(2 t ) =-4-2 tan(2 t ) ....
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