Prove of product rule and chain rule

3.2.1 Prove of product rule and chain rule

We discuss the derivative of the product function \(h(x) = f(x) g(x)\). So we find \begin{eqnarray*}h(x)-h(y) & = & f(x) g(x) - f(y) g(y)\\ & = & f(x) g(x) - f(x) g(y) - f(y) g(y) + f(x) g(y)\\ & = & f(x) \left(g(x) - g(y)\right) + g(y) \left( f(x) - f(y) \right) \end{eqnarray*}

and finally \begin{eqnarray*}\lim_{x\rightarrow y}\frac{h(x)-h(y)}{x-y} & = & \lim_{x\rightarrow y} f(x) \frac{g(x) - g(y)}{x-y} + \lim_{x\rightarrow y} g(y) \frac{f(x) - f(y)}{x-y}\\ & = & f(x) \frac{dg}{dx}(x) + g(x) \frac{df}{dx}(x)\end{eqnarray*}

Second we discuss the derivative of \(h(t) = g(f(t))\). According to the definition of the derivative two functions \(u(t)\) and \(v(t)\) exist with \(u(0) = v(0) = 0\) and \begin{eqnarray*}f(t)-f(x) & = & \left( t-x \right) \left[\frac{df}{dx}(x) + u(t) \right]\\ g(s)-g(y) & = & \left( s-y \right) \left[\frac{dg}{dy}(y) + v(s) \right] \end{eqnarray*}

so \begin{eqnarray*}h(t)-h(x) & = & g(f(t))-g(f(x)) \\ & = & \left[f(t)-f(x) \right] \left[\frac{dg}{dy}(y) + v(s) \right]\\ & = & \left(t-x\right) \left[\frac{df}{dx}(x) + u(t) \right] \left[\frac{dg}{dy}(y) + v(s) \right] \end{eqnarray*}

and finally

\[\frac{dh}{dx}= \lim_{t \rightarrow x} \frac{ h(t) -h(x)}{t-x} = \frac{df}{dx}(x) \frac{dg}{dx}(f(x)) \]

With frame

Calculation rules for derivatives