Product Rule Derivative using First Principle of Derivatives

The Mathematician — Wed, 10 May 2023 13:00:08 +0000

How to prove the product rule derivative using first principle of derivatives

We will prove the product rule by the first principle of derivatives, the definition of the derivative. In other words, we will prove the next equality holds:

\begin{equation*} (fg)'(x) = f'(x)g(x) + f(x)g'(x). \end{equation*}

Proof of product rule derivative using first principle of derivatives

\begin{align*} (fg)'(x) &= \lim_{h \rightarrow 0} \frac{f(x + h)g(x + h) - f(x)g(x)}{h} \\ \end{align*}

We will implement in the numerator the next equality because it equals zero and therefore won’t change:

\begin{equation*} f(x)g(x + h) - f(x)g(x + h) = 0. \end{equation*}

So we get

\begin{align*} &\lim_{h \rightarrow 0} \frac{f(x + h)g(x + h) - f(x)g(x)}{h} = \\ &\lim_{h \rightarrow 0} \frac{f(x + h)g(x + h) - f(x)g(x + h) + f(x)g(x + h) - f(x)g(x)}{h} = \\ &\lim_{h \rightarrow 0} \frac{(f(x + h)-f(x))g(x + h) + f(x)(g(x + h) - g(x))}{h} = \\ & \lim_{h \rightarrow 0} \frac{f(x + h)-f(x)}{h}g(x + h) + f(x) \lim_{h \rightarrow 0} \frac{(g(x + h) - g(x))}{h} = \\ & f'(x)g(x) + f(x)g'(x) \end{align*}

because

\begin{equation*} \lim_{h \rightarrow 0} \frac{f(x + h)-f(x)}{h} = f'(x) \quad \text{and} \quad \lim_{h \rightarrow 0} f(x) \frac{(g(x + h) - g(x))}{h} = g'(x). \end{equation*}

Therefore

(fg)'(x) = f'(x)g(x) + f(x)g'(x)

, which completes the proof.

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Product Rule Derivative using First Principle of Derivatives

How to prove the product rule derivative using first principle of derivatives

Proof of product rule derivative using first principle of derivatives