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.
 A: 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
$$=\lim_{h \to 0} \frac{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}{h}\frac{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}$$
$$=\lim_{h \to 0} \frac{h (h + 2 x)}{h(\sqrt{(x+h)^2+9}+\sqrt{x^2+9})}$$
$$=\lim_{h \to 0} \frac{h + 2 x}{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}$$
$$=\frac{2 x}{\sqrt{x^2+9}+\sqrt{x^2+9}}$$
$$=\frac{2 x}{2\sqrt{x^2+9}}$$
$$={x\over {\sqrt{x^2+9}}}$$
A: $lim_{h\rightarrow 0}{{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}\over h}$
$=lim_{h\rightarrow 0}{{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}\over h}{{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}\over {\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}} $
$=lim_{h\rightarrow 0}{{2x+h}\over {\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}}$
$={x\over {\sqrt{x^2+9}}}$.
A: 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 $\frac{2x}{2\sqrt{x^2 +9}}= \frac{x}{\sqrt{x^2 +9}}$.
A: We have: $\sqrt{x^{2}+9}$
Using the definition of a derivative:
$\Rightarrow \dfrac{d}{dx}\big(\sqrt{x^{2}+9}\big)=\displaystyle\lim_{h\rightarrow{0}}\Bigg(\dfrac{\sqrt{(x+h)^{2}+9}-\sqrt{x^{2}+9}}{h}\Bigg)$
$\hspace{33 mm}=\displaystyle\lim_{h\rightarrow{0}}\Bigg(\dfrac{\sqrt{(x+h)^{2}+9}-\sqrt{x^{2}+9}}{h}\bullet{\dfrac{\sqrt{(x+h)^{2}+9}+\sqrt{x^{2}+9}}{\sqrt{(x+h)^{2}+9}+\sqrt{x^{2}+9}}}\Bigg)$
$\hspace{33 mm}=\displaystyle\lim_{h\rightarrow{0}}\Bigg(\dfrac{\big(\sqrt{(x+h)^{2}+9}\big)^{2}-\big(\sqrt{x^{2}+9}\big)^{2}}{h\hspace{1 mm}\big(\sqrt{(x+h)^{2}+9}+\sqrt{x^{2}+9}\big)}\Bigg)$
$\hspace{33 mm}=\displaystyle\lim_{h\rightarrow{0}}\Bigg(\dfrac{(x+h)^{2}+9-\big(x^{2}+9\big)}{h\hspace{1 mm}\big(\sqrt{(x+h)^{2}+9}+\sqrt{x^{2}+9}\big)}\Bigg)$
$\hspace{33 mm}=\displaystyle\lim_{h\rightarrow{0}}\Bigg(\dfrac{x^{2}+2hx+h^{2}+9-x^{2}-9}{h\hspace{1 mm}\big(\sqrt{(x+h)^{2}+9}+\sqrt{x^{2}+9}\big)}\Bigg)$
$\hspace{33 mm}=\displaystyle\lim_{h\rightarrow{0}}\Bigg(\dfrac{h\hspace{1 mm}\big(2x+h\big)}{h\hspace{1 mm}\big(\sqrt{(x+h)^{2}+9}+\sqrt{x^{2}+9}\big)}\Bigg)$
$\hspace{33 mm}=\displaystyle\lim_{h\rightarrow{0}}\Bigg(\dfrac{2x+h}{\sqrt{(x+h)^{2}+9}+\sqrt{x^{2}+9}}\Bigg)$
$\hspace{33 mm}=\dfrac{2x+0}{\sqrt{(x+0)^{2}+9}+\sqrt{x^{2}+9}}$
$\hspace{33 mm}=\dfrac{2x}{\sqrt{x^{2}+9}+\sqrt{x^{2}+9}}$
$\hspace{33 mm}=\dfrac{2x}{2\sqrt{x^{2}+9}}$
$\hspace{33 mm}=\dfrac{x}{\sqrt{x^{2}+9}}$
