`y = e^(-2t) cos(4t)` Find the derivative of the function.
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hkj1385
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Note:- 1) If y = e^(at) ; then dy/dt = a*e^(at) ; where 'a' = constant
2) If y = cos(at) ; then dy/dt = -a*sin(at) ; where 'a' = constant
3) If y = u*v ; where both 'u' & 'v' are functions of 'x' ; then by the multiplicity rule of differentiation ; dy/dx = y' = u*(dv/dx) + v*(du/dx)
Now, given y = {e^(-2t)}*cos(4t)
Thus, dy/dx = y' = -[2*{e^(-2t)}*cos(4t)] + [-4*sin(4t)*{e^(-2t)}]
or, dy/dx = y' = [-2*{e^(-2t)}]*[cos(4t) + 2sin(4t)]
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balajia | Student
`y=e^(-2t)cos(4t)`
`y'=(-4sin4t)e^(-2t)+cos4t.(-2)e^(-2t)`
`y'=-2(2sin4t.e^(-2t)+cos4t.e^(-2t))`
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