# Using the limit definition to find a derivative for $\frac{-5x}{2+\sqrt{x+3}}$

I am trying to find the derivative of $\frac{-5x}{2+\sqrt{x+3}}$ using the limit definition of a derivative.

$$\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$$

What I did is

$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}\dfrac{2-\sqrt{x+h+3}}{2-\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}\dfrac{2-\sqrt{x+3}}{2-\sqrt{x+3}}}{h}$$

But I am having problem finding the correct answer.

• Please include your working too so that we can correct your mistakes .Also is it $2$ or $3$? Jun 6 '18 at 19:02
• Is this an exercise from a book? Usually one would use the limit definition of "standard" functions to find derivatives (like of $\sqrt{x }$) and then use rules of derivatives (product rule, quotient rule.....) to find derivatives of your function. To use the limit definition here is not impossible, but there is so much algebra then... Jun 6 '18 at 19:54
• It might be easier if you rationalize the denominator first. (a few seconds later) On second thought, perhaps not! Jun 6 '18 at 20:07
• OK, I managed to get a "from first principles" calculation sketched out on paper --- after combining to a single fraction, split up into two fractions after cancelling $+10x$ and $-10x$ in the numerator (one fraction will have two numerator terms with $5x$ factors and the other fraction will have two numerator terms with an $h$ factor), and compute the limits of these two separately. One is easy (the one where the $h$'s cancel) and the other, after factoring out $5x$ divided by the product of the two radicals, you rationalize the numerator. I may write this up later, but don't have time now. Jun 6 '18 at 20:20
• What I thought I had to do now was canceled, so I'll type up a solution now. Jun 6 '18 at 20:31

You better not rush to multiply by conjugates: $$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}}{h}=\\ \lim_{h \to 0} \frac{\left(\dfrac{-5x}{2+\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}\right)+\dfrac{-5h}{2+\sqrt{x+h+3}}}{h}=\\ -5x\cdot \lim_{h \to 0} \frac{\dfrac{1}{2+\sqrt{x+h+3}}-\dfrac{1}{2+\sqrt{x+3}}}{h}-\lim_{h \to 0}\dfrac{5}{2+\sqrt{x+h+3}}=\\ -5x\cdot \lim_{h \to 0} \frac{\sqrt{x+3}-\sqrt{x+h+3}}{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})}-\dfrac{5}{2+\sqrt{x+3}}=\\ -5x\cdot \lim_{h \to 0} \frac{-h}{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})(\sqrt{x+3}+\sqrt{x+h+3})}-\dfrac{5}{2+\sqrt{x+3}}=\\ \frac{15x}{(2+\sqrt{x+3})^2\cdot 2\sqrt{x+3}}-\dfrac{5}{2+\sqrt{x+3}}=\\ \frac{5x+20\sqrt{x+3}+30}{(2+\sqrt{x+3})^2\cdot 2\sqrt{x+3}}.$$

Don't rationalize like that; instead consider \begin{align} &\lim_{h\to0}\frac{1}{h}\left( \frac{-5(x+h)}{2+\sqrt{x+h+3}}-\frac{-5x}{2+\sqrt{x+3}}\right) \\[6px] &\qquad= \lim_{h\to0}\frac{ -10x-5x\sqrt{x+3}-10h-5h\sqrt{x+3}+10x+5x\sqrt{x+h+3} }{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})} \\[6px] &\qquad= \lim_{h\to0}\frac{-10-5\sqrt{x+3}}{(2+\sqrt{x+h+3})(2+\sqrt{x+3})}+ \lim_{h\to0}\frac{5x(\sqrt{x+h+3}-\sqrt{x+3})}{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})} \\[6px] &\qquad=\frac{-10-5\sqrt{x+3}}{2(2+\sqrt{x+3})}+ \frac{5x}{2(2+\sqrt{x+3})}\lim_{h\to0}\frac{\sqrt{x+h+3}-\sqrt{x+3}}{h} \end{align} and you should be able to finish up.

The difference quotient to which we are to apply the operation $\;\lim\limits_{h \rightarrow 0}\;$ is

$$\frac{f(x+h) - f(x)}{h} \;\; = \;\; \left[\frac{-5(x+h)}{2 + \sqrt{(x+h) + 3\;}\;} \; - \; \frac{-5x}{2 + \sqrt{x+3\;}\;} \right] \; \div \; h$$

$$= \;\; \left[\frac{-5(x+h)\left(2+\sqrt{x+3\;}\right) \;\; - \;\; (-5x)\left(2 + \sqrt{x+h+3\;}\right)\;}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)}\right] \; \div \; h$$

$$= \;\; \frac{-10x - 5x\sqrt{x+3\;} - 10h - 5h\sqrt{x+3\;} + 10x + 5x\sqrt{x+h+3\;}\;}{h\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)}$$

$$= \;\; \frac{-5x\sqrt{x+3\;} - 10h - 5h\sqrt{x+3\;} + 5x\sqrt{x+h+3\;}\;}{h\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)}$$

$$= \;\; \frac{-5x\sqrt{x+3\;} \; + \; 5x\sqrt{x+h+3\;}\;}{h\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} \;\; + \;\; \frac{-10h \; - \; 5h\sqrt{x+3\;}\;}{h\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)}$$

$$= \;\; \frac{5x}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} \; \cdot \; \frac{\sqrt{x+h+3\;} \; - \; \sqrt{x+3\;}\;}{h}$$

$$+ \;\; \frac{-10 \; - \; 5\sqrt{x+3\;}\;}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)}$$

This last expression has the form $\;A \cdot B \; + \; C,\;$ where for $\;h \rightarrow 0\;$ we have

$$A \;\; = \;\; \frac{5x}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} \;\; \longrightarrow \;\; \frac{5x}{\left(2 + \sqrt{x+3\;}\right)^2}$$

and

$$B \;\; = \;\; \frac{\sqrt{x+h+3\;} \; - \; \sqrt{x+3\;}}{h} \;\; = \;\; \frac{\sqrt{x+h+3\;} \; - \; \sqrt{x+3\;}}{h} \cdot \frac{\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}}{\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}}$$

$$= \;\; \frac{(x+h+3) \; - \; (x+3)\;}{h\left(\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}\right)} \;\; = \;\; \frac{h}{h\left(\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}\right)}$$

$$= \;\; \frac{1}{\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}\;} \;\; \longrightarrow \; \frac{1}{2\sqrt{x+3\;}\;}$$

and

$$C \;\; = \;\; \frac{-10 \; - \; 5\sqrt{x+3\;}\;}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} \;\; \longrightarrow \;\; \frac{-10 \; - \; 5\sqrt{x+3\;}}{\left(2 + \sqrt{x+3\;}\right)^2}$$

Therefore, the derivative is

$$A \cdot B \; + \; C \;\; = \;\; \frac{5x}{\left(2 + \sqrt{x+3\;}\right)^2} \; \cdot \; \frac{1}{2\sqrt{x+3\;}\;} \;\; + \;\; \frac{-10 \; - \; 5\sqrt{x+3\;}}{\left(2 + \sqrt{x+3\;}\right)^2}$$

I'll leave to you the verification that this expression is equal to the expression you get by differentiating using short-cut rules.