The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)}{(x-a)}$
$0=(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$(x^2-a^2)^{\frac{1}{2}}(2a)=\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$2a\sqrt{x^2-a^2}=\frac{a(x+a)}{2\sqrt{x^2-a^2}}$
$2a(x+a)(x-a)=a(x+a)$
$2(x-a)=a$
$2x-2a=a$
$x=\frac{3}{2}a$
My book says x=2a. I'm not sure where I went wrong. Thanks in advance
 A: For the ease of computation I use $$x=a\sec2\theta=a\cdot\frac{1+t^2}{1-t^2}$$ where $t=\tan\theta$ and $0\le\theta\le\dfrac\pi2$
So, $f(t)=\dfrac{4a^3}{t(1-t^2)}$
We need to attain the maximum positive value of $g(t)=t-t^3$ which can be achieved using Second derivative test
A: Note
$$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}\color{#C00000}{\cdot 2x}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$$ 
And Not:
$$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$$ 

This is because:
$$\frac{d}{dx} {f(x)}^n = n (f(x))^{n-1}\color{green}{f'(x)}$$
So, 
$$\frac{d}{dx} (x^2-a^2)^{\frac{1}{2}}=\frac{1}{2} (x^2-a^2)^{-\frac{1}{2}} \color{green}{2x} $$
A: You differentiated $\sqrt{x^2-a^2}$ incorrectly. It looks like you forgot to apply the chain rule.
A: I assume the constant $a$ is positive. This is because if $a\lt 0$ there is no minimum.
We equivalently want to minimize $\frac{(x+a)^4}{x^2-a^2}$ with $x\gt a$, or equivalently minimize
$$\frac{(x+a)^3}{x-a}$$
with $x\gt a$. The derivative is
$$\frac{3(x+a)^2(x-a)-(x+a)^3}{(x-a)^2},$$
difficult to get wrong. The top is 
$$(x+a)^2[3(x-a)-(x+a)], \quad\text{which is}\quad (x+a)^2(2x-4a).$$
Remark: Under suitable conditions, maximizing/minimizing $f(x)$ is equivalent to maximizing $f^2(x)$. Replacing $f(x)$ by $f^2(x)$ is a frequently useful little trick.
