Differentiate $e^{a x} c o s (b x + c) .$

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Solution

Using the product rule, we have

$\frac{d}{d x} [e^{a x} c o s (b x + c) .]$

$e^{a x} . \frac{d}{d x} [c o s (b x + c)] + c o s (b x + c) . \frac{d}{d x} (e^{a x})$

$= e^{a x} [- s i n (b x + c)] . \frac{d}{d x} . (b x + c) + c o s (b x + c) . e^{a x} . \frac{d}{d x} (a x)$ [using the chain rule ]

$= b e^{a x} s i n (b x + c) + a^{a x} c o s ‘ (b x + c)$

$= e^{a x} [a c o s (b x + c) - b s i n (b x + c) .]$

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