One of my Calculus lecture slides has the following exercise on it:
Let $g(x) = \frac{2}{\sqrt{x}}$. Calculate $g'(4)$ directly from the definition of derivative as a limit.
Since we already learned the basic differentiation rules, I can tell that $\require{cancel} \cancel{g'(4) = 0}$ $g'(4)= – \frac{1}{8}$. But I don't know how to calculate this result from the definition of derivative as a limit.
We learned of two equivalent definitions of derivatives as a limit:
$$ f'(a) = \lim\limits_{x \to a} \frac{f(x) – f(a)}{x – a} \tag{1}\label{1}$$
and
$$ f'(a) = \lim\limits_{h \to 0}\frac{f(a + h) – f(a)}{h}, \text{ where } h = x – a \tag{2}\label{2}$$
Using \eqref{1}:
$$g'(4) = \lim\limits_{x \to 4} \frac{g(x) – g(4)}{x – 4} = \lim\limits_{x \to 4} \frac{\frac{2}{\sqrt{x}} – 1}{x – 4} = \lim\limits_{x \to 4} \frac{2x^{-1} – 1}{x – 4}$$
How do I proceed from here?
Alternately, using \eqref{2}:
$$ g'(4) = \lim\limits_{h \to 0} \frac{g(4 + h) – 1}{h} = \lim\limits_{h \to 0} \frac{2(4 + h)^{-1/2} – 1}{h}$$
How do I proceed from here?
Best Answer
$g'(4)=\lim \frac {2-\sqrt {4+h}} {h\sqrt {4+h}}=\lim \frac {4-(4+h)} {h (2+\sqrt {4+h})\sqrt {4+h}}=-\frac 1 8$.