differentiation

[red and white kop!]

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

 

[daon]

You can use the product rule:

$\displaystyle D_x[\frac{1}{x-6} \cdot \frac{1}{x-2}] = D_x[\frac{1}{x-6}](\frac{1}{x-2}) + (\frac{1}{x-6})D_x[\frac{1}{x-2}]$

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


Or, you may multiply it out and use the chain rule:

$\displaystyle D_x[\frac{1}{x-6} \cdot \frac{1}{x-2}] = D_x[(x^2-8x+12)^{-1}]$

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

 

[Deleted member 4993]

Another similar but different representation of the same way:

$\displaystyle D_x\left[\frac{1}{(x-6)(x-2)}\right ]$

$\displaystyle = \, \frac{1}{4} \cdot D_x\left[\frac{1}{(x-6)} - \frac{1}{(x-2)}\right ]$

$\displaystyle = \, \frac{1}{4} \cdot\left[\frac{1}{(x-2)^2} \, - \, \frac{1}{(x-6)^2}\right ]$

Simplify further if needed.....

 

[red and white kop!]

thanks people