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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!

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With the expression $f(x+h)-f(x)$, you could try to get a common denominator and then multiply by the conjugate, of the expression in the numerator, on the top and bottom. – mwmjp Aug 7 '14 at 19:58
You should start by writing down the expression for $\frac{1}{h}[f(x+h)-f(x)]$ and then simplifying it. Then take the limit as $h$ tends to $0$. – JimmyK4542 Aug 7 '14 at 20:01
Just for everyone's information, this is sometimes called "differentiation from first principles". – michaelb958 Aug 8 '14 at 6:22
up vote 6 down vote accepted

$$\eqalign{ \lim_{h \to 0} \frac{f\left(x+h\right)-f\left(x\right)}{h}&=\lim_{h \to 0} \frac{\frac{1}{\sqrt{1+\left(x+h\right)^2}}-\frac{1}{\sqrt{1+x^2}}}{h}\\ &=\lim_{h \to 0} \frac{\sqrt{1+x^2}-\sqrt{1+\left(x+h\right)^2}}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}}\\ &=\lim_{h \to 0} \frac{\left(\sqrt{1+x^2}-\sqrt{1+\left(x+h\right)^2}\right)\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\ &=\lim_{h \to 0} \frac{1+x^2-\left(1+\left(x+h\right)^2\right)}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\ &=\lim_{h \to 0} \frac{-2xh-h^2}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)}\\ &=\lim_{h \to 0} \frac{-2x-h}{\sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}\left(\sqrt{1+x^2}+\sqrt{1+\left(x+h\right)^2}\right)} \\ &=\frac{-2x}{\left(1+x^2\right) \cdot 2 \cdot \sqrt{1+x^2}} \\ &=-\frac{x}{\left(1+x^2\right)^{{3}/{2}}} . }$$

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