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Differentiate e^(ax) cos(bx + c)....

Differentiate `e^(ax) cos(bx + c)`.

Text Solution

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Using the product rule, we have
`d/(dx) [e^(ax) cos (bx + c)]`
`= e^(ax). d/(dx) [cos (bx + c)] + cos (bx + c) . (d)/(dx)(e^(ax))`
`= e^(ax) [ -sin(bx + c)]. (d)/(dx)(bx+c) + cos(bx + c).e^(ax).(d)/(dx) (ax)`
(using the chain rule)
`= -be^(ax) sin (bx + c) + ae^(ax) cos (bx + c)`.
`= e^(ax) [a cos(bx+ c) - b sin (bx + c)]`.
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