Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 38

Determine the derivative of the function $y = (ax + \sqrt{x^2+b^2})^{-2}$


$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} (ax + \sqrt{x^2+b^2})^{-2} && \text{(where } a \text{ and } b \text{ are constant)}\\
\\
y' &= -2 (ax + \sqrt{x^2+b^2})^{-3} \frac{d}{dx} \left[ ax + ( x^2+b^2)^{\frac{1}{2}} \right]\\
\\
y' &= -2 (ax + \sqrt{x^2+b^2})^{-3} \left[ a + \frac{1}{2} (x^2+b^2)^{\frac{-1}{2}} \frac{d}{dx} ( x^2 + b^2)\right]\\
\\
y' &= -2 (ax + \sqrt{x^2+b^2})^{-3} \left[ a + \frac{1}{\cancel{2}} (x^2+b^2)^{\frac{-1}{2}}(\cancel{2}x) \right]\\
\\
y' &= \left[ \frac{-2}{(ax+\sqrt{x^2+b^2})^3}\right] \left[ a + \frac{x}{(x^2+b^2)^{\frac{1}{2}}}\right]
\end{aligned}
\end{equation}
$