Finding the derivative using the definition? Calculate the derivate of the given function directly from the definition of derivative, and express the result using differentials
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
when $f(x)= 1/\sqrt{1+x^2}$
any tips/solutions on how to get started on this one? I am able to do more basic problems, but not with root etc! thanks for tips/advice/solutions!
 A: $$\eqalign{
\lim_{h \to 0} \frac{f\left(x+h\right)-f\left(x\right)}{h}&=\lim_{h \to 0} \frac{\frac{1}{\sqrt{1+\left(x+h\right)^2}}-\frac{1}{\sqrt{1+x^2}}}{h}\\
&=\lim_{h \to 0} \frac{\sqrt{1+x^2}-\sqrt{1+\left(x+h\right)^2}}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}}\\
&=\lim_{h \to 0} \frac{\left(\sqrt{1+x^2}-\sqrt{1+\left(x+h\right)^2}\right)\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\
&=\lim_{h \to 0} \frac{1+x^2-\left(1+\left(x+h\right)^2\right)}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\
&=\lim_{h \to 0} \frac{-2xh-h^2}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\
&=\lim_{h \to 0} \frac{-2x-h}{\sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)} \\
&=\frac{-2x}{\left(1+x^2\right) \cdot 2 \cdot \sqrt{1+x^2}} \\ 
&=-\frac{x}{\left(1+x^2\right)^{{3}/{2}}} .
}$$
