# Let $g(x) = \frac{2}{\sqrt{x}}$. Calculate $g'(4)$ directly from the definition of derivative as a limit.

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?

$$\dfrac{2-√x}{√x(x-4)}=\dfrac{(4 -x)}{(2+√x)√x(x-4)}$$

$$= -\dfrac{1}{√x(2+√x)};$$

Take the limit $$x \rightarrow 4$$.

$$g'(4)=\lim \frac {2-\sqrt {4+h}} {h\sqrt {4+h}}=\lim \frac {4-(4+h)} {h (2+\sqrt {4+h})\sqrt {4+h}}=-\frac 1 8$$.

• I think the answer is $\frac {-1} 8$ – J. W. Tanner May 31 '19 at 12:08

Hint: Use the formula $$\sqrt{a}-\sqrt{b} = {a-b\over \sqrt{a}+\sqrt{b} }$$

$$\lim_{h\to0}\dfrac{\ 2 { (4+h)^{-1/2}}-1}h=\lim_{h\to0}\dfrac{(1+\frac h4)^{-1/2}-1}h=\lim_{h\to0}\dfrac{(1-\frac12\frac h4+O(h^2))-1} h=-\dfrac {1}8$$