1. ## Check some derivatives please.

Hi:
Could somebody check this for me please.
thanks you

2. Hello stealthmaths
Originally Posted by stealthmaths
Hi:
Could somebody check this for me please.
thanks you

They look fine to me, except that in (i) you have found $\frac{d^2y}{dx^2}$.

3. ahh yes. I put the first derivative instead of the second.

ahh wait. I still need to find the second derivative of (ii)

4. Hi:
part iib)

5. Hello stealthmaths
Originally Posted by stealthmaths
Hi:
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 $3$ at the start, and then factorise where you can:

$\tfrac{1}{3}y= \frac{x+1}{x^2-9}$

$\Rightarrow \frac13\frac{dy}{dx}= \frac{(x^2-9)(1)-(x+1)(2x)}{(x^2-9)^2}$
$=\frac{-x^2-2x-9}{(x^2-9)^2}$
$\Rightarrow \frac13\frac{d^2y}{dx^2}= \frac{(x^2-9)^2(-2x-2)-(-x^2-2x-9)2(x^2-9)(2x)}{(x^2-9)^4}$
$=2\frac{(x^2-9)^2(-x-1)+(x^2+2x+9)(x^2-9)(2x)}{(x^2-9)^4}$

$=2\frac{(x^2-9)(-x-1)+(x^2+2x+9)(2x)}{(x^2-9)^3}$

$=2\frac{-x^3-x^2+9x+9+2x^3+4x^2+18x}{(x^2-9)^3}$

$\Rightarrow \frac{d^2y}{dx^2}=\frac{6(x^3+3x^2+27x+9)}{(x^2-9)^3}$

But I get things wrong sometimes, too, so check my working!

I'm confused. You divide by 3 and just take the 1/3 out of the problem (on the top line you gave me)?

7. Somehow I doubt you rarely make a mistake Grandad(not unintentionally anyway)
I have gone through the work and I think it looks fine - except I'm not sure about the last line where you have a 6 - should this not be a 2?
I have marked it in red:
Thank you

8. Hello stealthmaths

I'm sure you'll kick yourself. I began by dividing both sides of the equation by $3$ to keep the numbers easier - hence the $\tfrac13y$, the $\frac13\frac{dy}{dx}$ and the $\frac13\frac{d^2y}{dx^2}$.

So at the end, I've simply multiplied by $3$ to get back to what we want, namely $\frac{d^2y}{dx^2}$.