linearity of differentiation

Linearity of differentiation

The derivative of any linear combination of functions equals the same linear combination of the derivatives of the functions. Thus, for any functions \[f\] and \[g\], and \[a,b\in\mathbb{R}\],

  • The derivative of \[h(x)=af(x)+bg(x)\] is \[h^{\prime}(x)=af^{\prime}(x)+bg^{\prime}(x)\]
  • \[(af)^{\prime}=af^{\prime}\]
  • \[(f+g)^{\prime}=f^{\prime}+g^{\prime}\]
  • \[(f-g)^{\prime}=f^{\prime}-g^{\prime}\]

Linearity

Let \[a,b\in \mathbb{R}\] and \[f,g\] be functions. Let \[j(x)=af(x)+bg(x)\]. To prove \[j^{\prime}(x)=af^{\prime}(x)+bg^{\prime}(x)\]:

\begin{align*} j^{\prime}(x)&=\lim_{\Delta x\to0}\frac{j(x+\Delta x)-j(x)}{\Delta x}\\ &=\lim_{\Delta x\to0}\frac{af(x+\Delta x)+bg(x+\Delta x)-af(x)-bg(x)}{\Delta x}\\ &=\lim_{\Delta x\to0}\left( \frac{af(x+\Delta x)-af(x)}{\Delta x}+\frac{bg(x+\Delta x)-bg(x)}{\Delta x} \right)\\ &=af^{\prime}(x)+bg^{\prime}(x)\\ \end{align*}

\[(af)^{\prime}=af^{\prime}\]

Let \[a\in\mathbb{R}\] and \[f\] be a function. Let \[j(x)=af(x)\]:

\begin{align*} j^{\prime}(x)&=\lim_{\Delta x\to0}\frac{j(x+\Delta x)-j(x)}{\Delta x}\\ &=\lim_{\Delta x\to0}\frac{af(x+\Delta x)-af(x)}{\Delta x}\\ &=af^{\prime}(x)\\ \end{align*}

\[(f\pm g)^{\prime}=f^{\prime}\pm g^{\prime}\]

Let \[f,g\] be functions. Let \[j(x)=f(x)\pm g(x)\].

\begin{align*} j^{\prime}(x)&=\lim_{\Delta x\to0}\frac{j(x+\Delta x)-j(x)}{\Delta x}\\ &=\lim_{\Delta x\to0}\frac{f(x+\Delta x)\pm g(x+\Delta x)-f(x)\mp g(x)}{\Delta x}\\ &=f^{\prime}(x)\pm g^{\prime}(x)\\ \end{align*}
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