<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] So we have $f(x) = \sqrt{x^2 +9}$.
Then using the definition of the derivative, we have $f'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)^2 +9} - \sqrt{x^2+9}}{h} =\lim_{h \to 0} \frac{\sqrt{(x+h)^2 +9} - \sqrt{x^2+9}}{h} * \frac{\sqrt{(x+h)^2 +9}+ \sqrt{x^2 +9}}{\sqrt{(x+h)^2 +9} + \sqrt{x^2+9}} = \lim_{h \to 0} \frac{2xh + h^2}{h(\sqrt{(x+h)^2 +9} + \sqrt{x^2 +9})}= \lim_{h \to 0} \frac{2x + h}{\sqrt{(x+h)^2 +9} + \sqrt{x^2 +9}}$.
As $h \to 0$, the limit tends to $\frac{2x}{2\sqrt{x^2 +9}}= \frac{x}{\sqrt{x^2 +9}}$.
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