<s>[INST] How do you find the derivative of $\sqrt{x^2+9}$ using the definition of a derivative? I had this question on a not-so-recent math test. It asked to find the derivative of $\sqrt{x^2+9}$ with the definition of a derivative. Using the chain rule, I can figure out that the derivative is $\frac x{\sqrt{x^2+9}}$, but how can it be done with only the definition of a derivative? I tried multiplying both sides of the fraction by the square roots, but that just makes a mess of everything.
 [/INST] By the definition of the derivative, what we want to evaluate is
$$\lim_{h \to 0} \frac{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}{h}$$
We now multiply the top and bottom by the conjugate of the numerator (to simplify the square roots) and simplify to get
$$=\lim_{h \to 0} \frac{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}{h}\frac{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}$$
$$=\lim_{h \to 0} \frac{h (h + 2 x)}{h(\sqrt{(x+h)^2+9}+\sqrt{x^2+9})}$$
$$=\lim_{h \to 0} \frac{h + 2 x}{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}$$
$$=\frac{2 x}{\sqrt{x^2+9}+\sqrt{x^2+9}}$$
$$=\frac{2 x}{2\sqrt{x^2+9}}$$
$$={x\over {\sqrt{x^2+9}}}$$
</s>