Linearity of the Derivative

Prerequisites:

The derivative is linear. Which makes sense if you think in terms of rates of change. Linearity means the rate of change of of a sum is the sum of the rates of change. Symbolically

ddx(f(x)+g(x))=ddxf(x)+ddxg(x)\frac{d}{dx}\left(f(x) + g(x)\right) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)

The second bit of linearity is that you can bring the constant outside

ddxcf(x)=cddxf(x)\frac{d}{dx} cf(x) = c \frac{d}{dx} f(x)

The important thing is this making sense intuitively, the proof isn't important but I'll present it for completeness. The trick is using the fact that limits are linear

(cf)′=lim⁡h→0cf(x+h)−cf(x)h=clim⁡h→0f(x+h)−f(x)h=cf′(x)\begin{aligned} (cf)' &= \lim_{h \to 0} \frac{c f(x+h) - c f(x)}{h} \\ &= c\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= cf'(x) \end{aligned}

And for the sum case (don't be afraid! it's just algebra!)

(f+g)′(x)=lim⁡h→0(f+g)(x+h)−(f+g)(x)h=lim⁡h→0f(x+h)+g(x+h)−f(x)−g(x)h=lim⁡h→0f(x+h)−f(x)h+lim⁡h→0g(x+h)−g(x)h=f′(x)+g′(x)\begin{aligned} (f + g)'(x) &= \lim_{h \to 0} \frac{(f + g)(x+h) - (f + g)(x)}{h} \\ &= \lim_{h \to 0} \frac{f(x+h) + g(x+h) - f(x) - g(x)}{h} \\ &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} + \lim_{h \to 0} \frac{g(x+h)-g(x)}{h} \\ &= f'(x) + g'(x) \end{aligned}