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An alternative notation for $\frac{d y}{d x}$ is $f^{'} (x) .$ So

\begin{aligned} y & = f (x) \\ \Rightarrow \frac{d y}{d x} & = f^{'} (x) \end{aligned}

( $f^{'} (x)$ is pronounced " $f$ prime of $x$ ")

A function $f$ is defined by
$f (x) = \sqrt{x}, x > 0$
Use first principles to find the gradient function $f^{'} (x) .$
Hence, find the value of $f^{'} (1)$ and explain the meaning of this value.

For (a), you should start with the definition

\begin{aligned} f^{'} (x) & = \frac{f (x + h) - f (x)}{h} \\ = \frac{\sqrt{x + h} - \sqrt{x}}{h} \end{aligned}

In order to proceed, trying multiplying the top and bottom of this fraction by

(\sqrt{x + h} + \sqrt{x})

The gradient of the curve at the point where $x = a$ is $\frac{1}{4}$ . Give the value of $a$ .