To find the derivative of the given function $g(x) = \sqrt{1 - x}$ using the definition of the derivative, we'll follow these steps.

Definition of Derivative:

The derivative of a function $g(x)$ at a point $x$ is given by the limit: $g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}$

Step-by-Step Process:

1. Substitute $g(x+h)$: $g(x+h) = \sqrt{1 - (x+h)} = \sqrt{1 - x - h}$

2. Substitute into the definition of derivative: $g'(x) = \lim_{h \to 0} \frac{\sqrt{1 - x - h} - \sqrt{1 - x}}{h}$

3. Rationalize the numerator: To simplify this expression, we multiply the numerator and denominator by the conjugate of the numerator: $\frac{\sqrt{1 - x - h} - \sqrt{1 - x}}{h} \times \frac{\sqrt{1 - x - h} + \sqrt{1 - x}}{\sqrt{1 - x - h} + \sqrt{1 - x}}$ This results in: $\frac{(1 - x - h) - (1 - x)}{h \left(\sqrt{1 - x - h} + \sqrt{1 - x}\right)}$ Simplifying the numerator: $\frac{-h}{h \left(\sqrt{1 - x - h} + \sqrt{1 - x}\right)}$ Cancel the $h$ terms: $\frac{-1}{\sqrt{1 - x - h} + \sqrt{1 - x}}$

4. Take the limit as $h \to 0$: As $h \to 0$, $\sqrt{1 - x - h} \to \sqrt{1 - x}$. Therefore, the limit becomes: $g'(x) = \frac{-1}{2\sqrt{1 - x}}$

Final Answer:

$g'(x) = \frac{-1}{2\sqrt{1 - x}}$

Would you like any additional details or have any questions about this solution?

Related Questions:

1. How do you use the product rule to differentiate complex functions?
2. What is the chain rule in calculus, and how does it apply to this problem?
3. How can you find the derivative of implicit functions?
4. What are some other methods to compute derivatives besides the definition of the derivative?
5. How do higher-order derivatives work, and how would you compute them for this function?

Tip:

Always check if you can simplify expressions by multiplying by conjugates when dealing with square roots in derivatives!