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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.