1. ## differentiation

how do i get the gradient function of 1/((x-6)(x-2))? where does the power of -1 come in?

2. Hi!

$\frac{1}{(x-6)(x-2)}=\frac{1}{x^{2}-8x+12}$

$\frac{d}{dx}\left(\frac{1}{x^{2}-8x+12}\right) = -\frac{(2x-8)}{(x^{2}-8x+12)^{2}}$

The negative sign comes from $\frac{1}{x^{2}-8x+12} = (x^{2}-8x+12)^{-1}$

Then you multiply by the "inner function".

3. can you show all your work please? and how am i supposed to work from here to get the turning points? i sense a tedious fourth-degree equation...

4. Hi!

$-\frac{(2x-8)}{(x^{2}-8x+12)^{2}}$ is the derivative.

Now factor out a 2.

$-2\cdot \frac{x-4}{x^{2}-8x+12}$ .

Note that the function is not defined when $x = 6 \mbox{ or } x=2$ (Why?)

Now from the factored expression for the derivative we see that the derivative is zero when $x=4$ .

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 $x=4$ must be a local maxima.

You also see that the derivative goes off to infinity as x approaches 2 and 6.

5. where did the power of two go in the beginning of this explanation?

6. Hi furor celtica
Originally Posted by furor celtica
how do i get the gradient function of 1/((x-6)(x-2))? where does the power of -1 come in?
As twig told you :
$\frac{d}{dx}\left(\frac{1}{x^{2}-8x+12}\right) = -\frac{(2x-8)}{(x^{2}-8x+12)^{2}}$

To find the gradient, just substitute the desired value of x to $\frac{dy}{dx}$.

Example : If you want to find the gradient at x= 1, subs. the value to $\frac{dy}{dx}$

Originally Posted by furor celtica
how am i supposed to work from here to get the turning points? i sense a tedious fourth-degree equation...
where did the power of two go in the beginning of this explanation?
To find turning points, set $\frac{dy}{dx}=0$. Then, you'll get :

$-\frac{(2x-8)}{(x^{2}-8x+12)^{2}}=0$

$2x-8=0$

$x=4$