How to prove that $\frac{d}{dx}\epsilon x^2=2\varepsilon x$ where $\varepsilon$ is just a constant How to prove that $\frac{d}{dx}\varepsilon x^2=2\varepsilon x$ where $\varepsilon$ is just a constant? Thanks in advance for your immense help.

If $f$ is differentiable, and $c$ is constant, $$\frac{c f(x+h)-cf(x)}h=c\frac{f(x+h)-f(x)}h$$ Hence $(cf)'=cf'$ by letting $h\to 0$.