Hello stealthmaths Originally Posted by
stealthmaths Hi:
Could somebody please check the last part of this question please?
part iib)
You have a sign wrong. See the attached - should be a minus sign at the point I've indicated.
I should set it out like this. First notice that it's easier to divide by $\displaystyle 3$ at the start, and then factorise where you can:
$\displaystyle \tfrac{1}{3}y= \frac{x+1}{x^2-9}$
$\displaystyle \Rightarrow \frac13\frac{dy}{dx}= \frac{(x^2-9)(1)-(x+1)(2x)}{(x^2-9)^2}$$\displaystyle =\frac{-x^2-2x-9}{(x^2-9)^2}$
$\displaystyle \Rightarrow \frac13\frac{d^2y}{dx^2}= \frac{(x^2-9)^2(-2x-2)-(-x^2-2x-9)2(x^2-9)(2x)}{(x^2-9)^4}$$\displaystyle =2\frac{(x^2-9)^2(-x-1)+(x^2+2x+9)(x^2-9)(2x)}{(x^2-9)^4}$
$\displaystyle =2\frac{(x^2-9)(-x-1)+(x^2+2x+9)(2x)}{(x^2-9)^3}$
$\displaystyle =2\frac{-x^3-x^2+9x+9+2x^3+4x^2+18x}{(x^2-9)^3}$
$\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\frac{6(x^3+3x^2+27x+9)}{(x^2-9)^3}$
But I get things wrong sometimes, too, so check my working!
Grandad