The derivative of e^3x sin4x with respect to c is ,
Ps : the answer is
5e^3x sin( 4x + tan^-1 (4/3 ))
I need the process please !!
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Answer:
see below
Step-by-step explanation:
d/dx [ e^3x sin4x ]
= d/dx [ e^3x ] sin 4x + d/dx [ sin 4x ] e^3x
= 3 e^3x sin 4x + 4 cos 4x e^3x
= e^3x [ 3 sin 4x + 4 cos 4x ]
= e^3x 5 [ 3/5 sin 4x + 4/5 cos 4x ] .....(1)
now let ....
cos @ = 3/5 & sin @ = 4/5
then....
sin@/cos@ =4/3 ;tan @ =4/3; @= tan^-1 4/3
from (1).......
= e^3x 5 [ cos @ sin 4x + sin @ cos 4x ]
= 5 e^3x sin ( 4x + @ )
= 5 e^3x sin [ 4x + tan^-1 (4/3) ]
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