differentiate

We could use the Quotient Rule here......however....I generally like to use the Product/Power/Chain Rules, whenever possible....

Note that  √x^2+1/(x-9)^2   =  (x^2 + 1)^(1/2) * (x - 9)^(-2)

And taking the derivaive, we have

(1/2)(2x)(x^2 + 1)^(-1/2)* (x - 9)^(-2) +  (x^2 + 1)^(1/2)* (-2)(x - 9)^-3    (simplify)

x(x^2 + 1)^(-1/2)*(x - 9)^(-2)  - 2 (x^2 + 1)^(1/2)* (x - 9)^-3    (factor)

[(x^2 + 1)^(-1/2) * (x - 9)^(-3) ] * [ x(x-9) - 2(x^2 + 1)] =

[(x^2 + 1)^(-1/2) * (x - 9)^(-3) ] * [ x^2 - 9x - 2x^2 -2] =

[(x^2 + 1)^(-1/2) * (x - 9)^(-3) ] * [ -(x^2 + 9x + 2)]

And this is the same result that Melody and Radix found......!!!

Which bit is under the square root?

only the X^2+1 is under the square root

$$\\=\frac{\sqrt{x^2+1}}{(x-9)^2}\\\\
\frac{(x^2+1)^{1/2}}{(x-9)^2}\\\\
Let\\
u=(x^2+1)^{1/2}\qquad \qquad \qquad \qquad\qquad v=(x-9)^2\\\\
u'=(1/2)(x^2+1)^{-1/2}*2x \qquad \qquad v'=2(x-9)\\\\
u'=(x^2+1)^{-1/2}*x \qquad \qquad \qquad \;\;\; v'=2(x-9)\\\\
$Quotient Rule$\\\\
\boxed{y'=\frac{vu'-uv'}{v^2}}\\\\
y'=\frac{(x-9)^2*(x^2+1)^{-1/2}*x-(x^2+1)^{1/2}*2(x-9)}{(x-9)^4}\\\\
y'=\frac{(x-9)*(x^2+1)^{-1/2}*x-(x^2+1)^{1/2}*2}{(x-9)^3}\\\\
y'=\left(\frac{x(x-9)}{(x^2+1)^{1/2}}-\frac{2(x^2+1)}{(x^2+1)^{1/2}}\right)\div (x-9)^3\\\\$$

$$\\y'=\left(\frac{x(x-9)-2(x^2+1)}{(x^2+1)^{1/2}}\right)\div (x-9)^3\\\\
y'=\left(\frac{x^2-9x-2x^2-2}{\sqrt{x^2+1}}\right)\div (x-9)^3\\\\
y'=\frac{-x^2-9x-2}{(x-9)^3\sqrt{x^2+1}} \\\\$$

Now, if it is not riddled with careless errors (which it probably is) that is your answer :))

Thanks Radix I have now repaired mine and it is the same as yours.  That is a good sign LOL   :))

Hello Melody,

I have this result:

$${\frac{\left({\mathtt{\,-\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{\left({\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\sqrt{{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}\right)}}$$

Gruß radix !

Thanks Radix, I have repaired mine now.  :))