Show that Dn/dxn(eax sin bx) = rn eax sin (bx + n) Where a and b are
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Dn/dxn(eax sin bx) = rn eax sin (bx + nθ)
Where a and b are positive numbers, r2 = a2 + b2, and θ = tan-1(b/a).
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Consider the statement that d n dx n e nx sin bx r n e a...View the full answer
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