Mathematics · Ch 18 — Differentiation

Theorem 1 — Derivative of Sum of Functions

18.2.1

Theorem 1 — Derivative of Sum of Functions

Theorem 1. If $u$ and $v$ are differentiable functions of $x$ such that $y=u+v$ , then

\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}

Proof. Given $y=u+v$ where $u,v$ are differentiable functions of $x$ . Let $x$ receive a small increment $\delta x$ ; correspondingly $u$ becomes $u+\delta u$ , $v$ becomes $v+\delta v$ , and $y$ becomes $y+\delta y$ . Then

y+\delta y=(u+\delta u)+(v+\delta v)

Subtracting $y=u+v$ :

\delta y=(u+\delta u)+(v+\delta v)-(u+v)=\delta u+\delta v

Since $\delta x\ne0$ , divide throughout by $\delta x$ :

\dfrac{\delta y}{\delta x}=\dfrac{\delta u}{\delta x}+\dfrac{\delta v}{\delta x}

Taking the limit as $\delta x\to0$ :

\lim_{\delta x\to0}\dfrac{\delta y}{\delta x}=\lim_{\delta x\to0}\dfrac{\delta u}{\delta x}+\lim_{\delta x\to0}\dfrac{\delta v}{\delta x}\qquad\ldots(\mathrm I)

Since $u$ and $v$ are differentiable functions of $x$ , …

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