How do you use the definition of a derivative to find f' given #f(x)=sqrt(4x+3)# at x>-3/4?

To find the derivative (f'(x)) of the function (f(x) = \sqrt{4x + 3}) using the definition of a derivative, we use the formula:

[f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}]

First, we substitute the function into the formula:

[f'(x) = \lim_{h \to 0} \frac{\sqrt{4(x + h) + 3} - \sqrt{4x + 3}}{h}]

Next, we rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator:

[f'(x) = \lim_{h \to 0} \frac{\sqrt{4(x + h) + 3} - \sqrt{4x + 3}}{h} \cdot \frac{\sqrt{4(x + h) + 3} + \sqrt{4x + 3}}{\sqrt{4(x + h) + 3} + \sqrt{4x + 3}}]

[= \lim_{h \to 0} \frac{(4(x + h) + 3) - (4x + 3)}{h(\sqrt{4(x + h) + 3} + \sqrt{4x + 3})}]

[= \lim_{h \to 0} \frac{4x + 4h + 3 - 4x - 3}{h(\sqrt{4(x + h) + 3} + \sqrt{4x + 3})}]

[= \lim_{h \to 0} \frac{4h}{h(\sqrt{4(x + h) + 3} + \sqrt{4x + 3})}]

[= \lim_{h \to 0} \frac{4}{\sqrt{4(x + h) + 3} + \sqrt{4x + 3}}]

[= \frac{4}{\sqrt{4x + 3} + \sqrt{4x + 3}}]

[= \frac{4}{2\sqrt{4x + 3}}]

[= \frac{2}{\sqrt{4x + 3}}]

Thus, (f'(x) = \frac{2}{\sqrt{4x + 3}}) when (x > -\frac{3}{4}).