# Thread: derivate from first principles.

1. ## derivate from first principles.

Need help with derivate of this function.

f(x) = 1/x, x can not be 0

lim (f(x+h)-f(x)) / h =
h->0

lim ((1/(x-h)) - (1/x)) / h =
h->0

lim (x-(x+h)) / (x(x+h)h) =
h->0

lim (-h) / (hx^2 + xh^2) =
h->0

Need help with this step

-1/x^2

2. Factor out h from the denominator and cancel (it only approaches 0 so it can cancel)

$\displaystyle \displaystyle \lim_{h \to 0} \left(\dfrac{-h}{xh(x+h)}\right) = \lim_{h \to 0} \left(-\dfrac{1}{x(x+h)}\right)$

Since we are taking the limit of h, not x it follows that $\displaystyle \displaystyle \lim_{h \to 0} (x+h) = x$. When this is put back into the original equation you get:

$\displaystyle -\dfrac{1}{x(x)} = -\dfrac{1}{x^2}$

Which is what we'd expect knowing the power rule of differentiation.

3. Originally Posted by paulaa
Need help with derivate of this function.

f(x) = 1/x, x can not be 0

lim (f(x+h)-f(x)) / h =
h->0

lim ((1/(x-h)) - (1/x)) / h =
h->0

lim (x-(x+h)) / (x(x+h)h) =
h->0

lim (-h) / (hx^2 + xh^2) =
h->0

Need help with this step

-1/x^2
When you are at this step you are almost done
$\displaystyle \displaystyle \lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h} = \lim_{h \to 0} \frac{\frac{x-(x+h)}{(x)(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{x(x+h)h} = \lim_{h \to 0}\frac{-1}{x(x+h)}=-\frac{1}{x^2}$

too slow