Question

Use the definition of derivative to find the derivative of h(t) = √ 2x − 1.

Answer 1

Definition of derivative is:

$$h'(x) = \frac{h(x+p)-h(x)}{p}$$

where: $p= \Delta x$

Let $y=h(x)=\sqrt{2x − 1}$ , then

$$y+\Delta y=\sqrt{2(x+\Delta x) − 1}\\ y+\Delta y=\sqrt{2x+2\Delta x − 1}\\ \Delta y=\sqrt{2x+2\Delta x − 1}-y\\ \Delta y=\sqrt{2x+2\Delta x − 1}-\sqrt{2x-1}\\ \frac{\Delta y}{\Delta x}= \frac{\sqrt{2x+2\Delta x − 1}-\sqrt{2x-1}}{\Delta x}$$

Rationalising the numerator, we have:

$$\frac{\Delta y}{\Delta x}= \frac{\sqrt{2x+2\Delta x − 1}-\sqrt{2x-1}}{\Delta x} \times \frac{\sqrt{2x+2\Delta x − 1}+\sqrt{2x-1}}{\sqrt{2x+2\Delta x − 1}+\sqrt{2x-1}}\\ \frac{\Delta y}{\Delta x}= \frac{2x+2\Delta x − 1-(2x-1)}{\Delta x(\sqrt{2x+2\Delta x − 1}+\sqrt{2x-1})} \\ \frac{\Delta y}{\Delta x}= \frac{2x+2\Delta x − 1-2x+1}{\Delta x(\sqrt{2x+2\Delta x − 1}+\sqrt{2x-1})} \\ \frac{\Delta y}{\Delta x}= \frac{2\Delta x}{\Delta x(\sqrt{2x+2\Delta x − 1}+\sqrt{2x-1})} \\ \frac{\Delta y}{\Delta x}= \frac{2}{\sqrt{2x+2\Delta x − 1}+\sqrt{2x-1}} \\$$

Taking limits of both sides

$$\lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x}= \lim_{\Delta x \rightarrow 0}\frac{2}{\sqrt{2x+2\Delta x − 1}+\sqrt{2x-1}} \\ \frac{dy}{dx} =\frac{2}{\sqrt{2x − 1}+\sqrt{2x-1}} \\ \frac{dy}{dx} =\frac{2}{2\sqrt{2x-1}} \\ \frac{dy}{dx} =\frac{1}{\sqrt{2x-1}} \\$$

Thus:

$$h'(x) = \frac{dy}{dx} =\frac{1}{\sqrt{2x-1}} \\$$