Example Find the derivative of the following function．

$$\displaystyle{y = (e^{2x} \sin{3x})^2}$$

Answer To find the derivative of a composite function, it is important to know the followings:

1.	The derivative of a sum is the sum of the derivatives． $(f(x) + g(x))' = f'(x) + g'(x)$
2.	The derivative of a constant multiple is the constant times the derivative. $(\alpha f(x))' = \alpha f'(x)$
3.	Product Rule． $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$
4.	Quotient Rule. $(\frac{f(x)}{g(x)})' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$
5.	Derivative of Inverse function. $y = f^{-1}{x} \Leftrightarrow x = f(y) \Leftrightarrow 1 = \frac{df(y)}{dx} = \frac{df(y)}{dy}\frac{dy}{dx}$.
	$y = \sin^{-1}{x} \Leftrightarrow x = \sin{y} \Leftrightarrow 1 = \frac{d(\sin{y})}{dx} \Leftrightarrow 1 = \cos{y}\frac{dy}{dx}$
	$\Leftrightarrow \frac{dy}{dx} = \frac{1}{\cos{y}} = \frac{1}{\sqrt{1-x^2}}$ $(\cos{y} = \sqrt{1 - \sin^{2}{y}})$
6.	The derivative of a composite function． $(f(g(x))' = f'(g(x))g'(x)$
7.	Basic Formula for derivatives. $(e^{x})' = e^{x}, (\tan{x})' = \sec^{2}{x}, (\sec{x})' = \sec{x}\tan{x}$

We apply these rulles to the above problem.

$$((e^{2x} \sin{3x})^2)' = 2(e^{2x}\sin{3x})(e^{2x}\sin{3x})'$$

$$= 2(e^{2x}\sin{3x})[(e^{2x})'\sin{3x} + e^{2x}(\sin{3x})']$$

$$= 2e^{2x}\sin{3x}[2e^{2x}\sin{3x}+ 3e^{2x}\cos{3x}]$$

$$= 2e^{4x}\left(2\sin^{2}{3x} + 3\sin{3x}\cos{3x}\right)$$