Find the derivative of 1 / square root of 2x
from first principles.
hellp !
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when you see "from first principles" they are talking about using the definition of the derivative.
there are two (equivalent) definitions:
$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$
if this is the definition you have been taught, then you are required to find the limit
$\displaystyle f'(x) = \lim_{h \to 0} \frac {\frac 1{\sqrt{2(x + h)}} - \frac 1{\sqrt{2x}}}h$
alternatively, you may use the definition
$\displaystyle f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$
in this case you are required to find the limit
$\displaystyle f'(a) = \lim_{x \to a} \frac { \frac 1{\sqrt{2x}} - \frac 1{\sqrt{2a}}}{x - a}$
(and replace $\displaystyle a$ with $\displaystyle x$ when finished)
Notation: $\displaystyle f'(x)$ and $\displaystyle f'(a)$ mean "the derivative of $\displaystyle f(x)$ (with respect to $\displaystyle x$)" and "the derivative of $\displaystyle f(x)$ (with respect to x) at the point $\displaystyle x = a$" respectively.
here of course, $\displaystyle f(x) = \frac 1{\sqrt{2x}}$
$\displaystyle \frac {\frac 1{\sqrt{2(x + h)}} - \frac 1{\sqrt{2x}}}h = \frac {\frac {\sqrt{2x} - \sqrt{2(x + h)}}{\sqrt{2(x + h)} \sqrt{2x}}}h = \frac {\sqrt{2x} - \sqrt{2(x + h)}}{h( \sqrt{2(x + h)} \sqrt{2x})} = \frac {\sqrt{2x} - \sqrt{2(x + h)}}{h(\sqrt{4x(x + h)})}$