Try this for n = 1 first.

The derivative of e\^(ax) = ae\^(ax), and the derivative of sin(bx) is b(cos(bx)) by the chain rule. Now, use the product rule to find the (first) derivative of e\^(ax)(sin(bx)). Recall that (fg)' = f'g + fg'. Thus, the first derivative is (ae\^(ax))sin(bx) + (e\^(ax))(b(cos(bx)). Factoring out the e\^(ax) yields e\^(ax) times (asin(bx) + bcos(bx)). It is thus sufficient to verify that rsin(bx + tan\^{-1}(b/a)) = (asin(bx) + bcos(bx)).

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By the formula sin(a + b) = sin(a)cos(b) + cos(a)sin(b), we get that rsin(bx + tan\^{-1}(b/a)) = r(sin(bx)cos(tan\^{-1}(b/a))) + cos(bx)sin(tan(\^{-1}(b/a)))). As cos(tan\^{-1}(x)) = 1/sqrt(1 + x\^2) and sin(tan\^{-1}(x)) = x/sqrt(1 + x\^2), we get that r(sin(bx)cos(tan\^{-1}(b/a))) + cos(bx)sin(tan(\^{-1}(b/a)))) = r(sin(bx)/sqrt(1 + (b/a)\^2) + cos(bx)(b/a)/sqrt(1 + (b/a)\^2)). As r = sqrt(a\^2 + b\^2), you can do the simple algebra (by clearing denominators) to show that r(sin(bx)/sqrt(1 + (b/a)\^2) + cos(bx)(b/a)/sqrt(1 + (b/a)\^2)) = (asin(bx) + bcos(bx)) because I don't feel like writing it out.

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For the higher cases, you can check this inductively. Assume that the result holds for n, and show that it holds for n+1. In other words, you want to take (r\^n)(e\^(ax))(sin(bx + n theta)) and differentiate it with respect to x, then show that this equals (r\^(n+1))(e\^(ax))(sin(bx + (n+1) theta)) for an arbitrary (but fixed, and treated as a constant when differentiating) n. This is a simple exercise in using product rule and chain rule (using similar methods to the n=1 case!) so I'll let you handle that.