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