Calculus: Finding Arc Length--Squaring the Derivative Where did the -1/2 come from? Math Example about finding the arc length.
I have gotten the derivative of the equation. 
Here is the equation. 
$$f(x)=\frac{x^5}{5} + \frac{1}{12x^3}$$
Derivative of the equation is:
$$f'(x) = x^4 - \frac{1}{4x^4}$$
The next step is to square the derivative: 
$$f'(x)^2 = x^8 -\frac12 + \frac {1}{16x^8}$$
My question is where did $\left(-\frac 12\right)$ come from after squaring the derivative? 
Thank you. 
 A: Recall the identity
$$(a-b)^2=a^2-2ab+b^2.$$ 
Take $a=x^4$ and $b=\frac{1}{4x^4}$. We have $2ab=\frac{1}{2}$.
Remark: The fun part is that when you add $1$ (arclength formula), the $-\frac{1}{2}$ turns into $\frac{1}{2}$, and our new expression is $(a+b)^2$. So now we can take the square root, getting $a+b$, that is, $x^4+\frac{1}{4x^4}$. This is simple to integrate. 
The numbers in this arclength problem were carefully chosen to make the final integration easy. For most functions $y=y(x)$, even "nice" ones,  $\sqrt{1+\left(\frac{dy}{dx}\right)^2}$ does not have an elementary antiderivative. Because of that, the majority of textbook arclength problems are highly artificial.
A: In case, we don't remember formula then it can calculated by simple multiplication as follows $$f'(x)=x^4-\frac{1}{4x^4}$$  $$\implies (f'(x))^2=\left(x^4-\frac{1}{4x^4}\right)^2$$  $$=\left(x^4-\frac{1}{4x^4}\right)\left(x^4-\frac{1}{4x^4}\right)$$  $$=(x^4)(x^4)+\left(-\frac{1}{4x^4}\right)(x^4)+(x^4)\left(-\frac{1}{4x^4}\right)+\left(-\frac{1}{4x^4}\right)\left(-\frac{1}{4x^4}\right)$$ $$=x^8-\frac{1}{4}-\frac{1}{4}+\frac{1}{16x^8}$$ $$=x^8\color{blue}{-\frac{1}{2}}+\frac{1}{16x^8}$$ **
A: As Andre' Nicolas said, $(a- b)^2= a^2- 2ab+ b^2$
In particular, if $b= 1/4a$, $$(a- 1/4a)^2= a^2- 2a\frac{1}{4a}+ \frac{1}{(4a)^2}= a^2- \frac{1}{2}+ \frac{1}{16a^2}$$
