**Exercise:** Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.

- f(x)=x^2-2x^3
- g(x)=\sqrt{9-x}
- G(t)=\frac{1-2t}{3+t}
- f(x)=x^4

- \begin{aligned}f'(x) & = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ & = \lim_{h\to 0}\frac{(x+h)^2-2(x+h)^3-x^2+2x^3}{h} \\ & = \lim_{h\to 0}\frac{2hx+h^2-2(3hx^2+3h^2x+h^3)}{h} \\ & = \lim_{h\to 0}{2x+h-6x^2-6hx-2h^2} \\ & = 2x-6x^2\end{aligned}. The domains of both the function and its derivative are \mathbb{R}.
- \begin{aligned}g'(x) & = \lim_{h\to 0}\frac{g(x+h)-g(x)}{h} \\ & = \lim_{h\to 0}\frac{\sqrt{9-(x+h)}-\sqrt{9-x}}{h} \\ & = \lim_{h\to 0}\frac{-h}{h(\sqrt{9-(x+h)}+\sqrt{9-x})} \\ & = \lim_{h\to 0}\frac{-1}{\sqrt{9-(x+h)+\sqrt{9-x}}} \\ & = \frac{-1}{2\sqrt{9-x}}\end{aligned}. The domain of the function is (-\infty, 9]. While the domain of the derivative is (-\infty, 9).
- \begin{aligned}G'(t) & = \lim_{h\to 0}\frac{G(x+h)-G(x)}{h} \\ & = \lim_{h\to 0}\frac{\frac{1-2(t+h)}{3+(t+h)}-\frac{1-2t}{3+t}}{h} \\ & = \lim_{h\to 0}\frac{(1-2(t+h))(3+t)-(1-2t)(3+t+h)}{(3+t+h)(3+t)h} \\ & = \lim_{h\to 0}\frac{-7h}{(3+t+h)(3+t)h} \\ & = \frac{-7}{(3+t)^2}\end{aligned}. The domain of both the function and its derivative are (-\infty, -3)\cup(-3, \infty).
- \begin{aligned}f'(x) & = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ & = \lim_{h\to 0}\frac{(x+h)^4-x^4}{h} \\ & = \lim_{h\to 0}\frac{4x^3h+6x^2h^2+4xh^3+h^4}{h} \\ & = \lim_{h\to 0}{4x^3+6x^2h+4xh^2+h^3} \\ & = 4x^3\end{aligned}. The domains of both the function and its derivative are \mathbb{R}.