<s>[INST] 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. 
 [/INST] In case, we don't remember formula then it can calculated by simple multiplication as follows $$f'(x)=x^4-\frac{1}{4x^4}$$  $$\implies (f'(x))^2=\left(x^4-\frac{1}{4x^4}\right)^2$$  $$=\left(x^4-\frac{1}{4x^4}\right)\left(x^4-\frac{1}{4x^4}\right)$$  $$=(x^4)(x^4)+\left(-\frac{1}{4x^4}\right)(x^4)+(x^4)\left(-\frac{1}{4x^4}\right)+\left(-\frac{1}{4x^4}\right)\left(-\frac{1}{4x^4}\right)$$ $$=x^8-\frac{1}{4}-\frac{1}{4}+\frac{1}{16x^8}$$ $$=x^8\color{blue}{-\frac{1}{2}}+\frac{1}{16x^8}$$ **
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