how do i get the gradient function of 1/((x-6)(x-2))? where does the power of -1 come in?
Hi!
$\displaystyle \frac{1}{(x-6)(x-2)}=\frac{1}{x^{2}-8x+12} $
$\displaystyle \frac{d}{dx}\left(\frac{1}{x^{2}-8x+12}\right) = -\frac{(2x-8)}{(x^{2}-8x+12)^{2}} $
The negative sign comes from $\displaystyle \frac{1}{x^{2}-8x+12} = (x^{2}-8x+12)^{-1} $
Then you multiply by the "inner function".
Hi!
$\displaystyle -\frac{(2x-8)}{(x^{2}-8x+12)^{2}} $ is the derivative.
Now factor out a 2.
$\displaystyle -2\cdot \frac{x-4}{x^{2}-8x+12} $ .
Note that the function is not defined when $\displaystyle x = 6 \mbox{ or } x=2 $ (Why?)
Now from the factored expression for the derivative we see that the derivative is zero when $\displaystyle x=4 $ .
To determine if it is a local maxima or local minima, you could use the second derivative test, or simply make a table.
Note that if you plug in values slighly less than 4, you get a positive derivative, and values slight larger than 4 gives you a negative. So $\displaystyle x=4 $ must be a local maxima.
You also see that the derivative goes off to infinity as x approaches 2 and 6.
Hi furor celtica
As twig told you :
$\displaystyle \frac{d}{dx}\left(\frac{1}{x^{2}-8x+12}\right) = -\frac{(2x-8)}{(x^{2}-8x+12)^{2}}$
To find the gradient, just substitute the desired value of x to $\displaystyle \frac{dy}{dx}$.
Example : If you want to find the gradient at x= 1, subs. the value to $\displaystyle \frac{dy}{dx}$
To find turning points, set $\displaystyle \frac{dy}{dx}=0$. Then, you'll get :where did the power of two go in the beginning of this explanation?
$\displaystyle -\frac{(2x-8)}{(x^{2}-8x+12)^{2}}=0$
$\displaystyle 2x-8=0$
$\displaystyle x=4$