Sum and Product Rule
Useful rules for differentiating sums or products of differentiable functions.
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Study Sum and Product Rule
Now we consider rules for the derivatives of sums, products and quotients of functions.
Theorem 1. Summation Rule, Product Rule, Quotient Rule Let \(f,g: I\to\mathbb{R}\) be differentiable in \(x_0\in I\).
Then \(f+g\) is differentiable in \(x_0\) with \((f+g)'(x_0)=f'(x_0)+g'(x_0)\).
Then \(f\cdot g\) is differentiable in \(x_0\) with \((f\cdot g)'(x_0)=f'(x_0)\cdot g(x_0)+f(x_0)\cdot g'(x_0)\).
If \(g(x_0)\neq0\), then \(\frac{f}g(x_0)\) is differentiable in \(x_0\) with \[\left(\frac{f}g(x_0)\right)'(x_0)=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{g^2(x_0)}.\]
Proof: Let \(f(x)=f(x_0)+(x-x_0)\cdot\Delta_{f,x_0}(x)\), \(g(x)=g(x_0)+(x-x_0)\cdot\Delta_{g,x_0}(x)\). Then
\[(f+g)(x)=f(x)+g(x)=(f+g)(x_0)+(x-x_0)\cdot(\Delta_{f,x_0}(x)+\Delta_{g,x_0}(x)).\]
\[\begin{aligned} (f\cdot g)(x)=& (f(x_0)+(x-x_0)\cdot\Delta_{f,x_0}(x_0))(g(x_0)+(x-x_0)\cdot\Delta_{g,x_0}(x))\\ =&f(x_0)g(x_0)+(x-x_0)(\Delta_{f,x_0}(x_0)g(x_0)+\Delta_{g,x_0}(x)f(x_0))\\&\quad+(x-x_0)^2\Delta_{f,x_0}(x_0)\Delta_{g,x_0}(x) .\end{aligned}\] Thus, \[\begin{aligned} &\lim_{x\to x_0}\frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0}\\ =&\lim_{x\to x_0}\frac{(x-x_0)(\Delta_{f,x_0}(x)g(x_0)+\Delta_{g,x_0}(x)f(x_0))+(x-x_0)^2\Delta_{f,x_0}(x)\Delta_{g,x_0}(x)}{x-x_0}\\ =&\lim_{x\to x_0}\left((\Delta_{f,x_0}(x_0)g(x_0)+\Delta_{g,x_0}(x)f(x_0)+(x-x_0)\Delta_{f,x_0}(x)\Delta_{g,x_0}(x)\right)\\ =&\Delta_{f,x_0}(x)g(x_0)+\Delta_{g,x_0}(x)f(x_0) \end{aligned}\]
For convenience, we assume that \(f\equiv1\) (the general result follows by an application of the product rule). Then \[\frac1{g(x)}-\frac1{g(x_0)}=\frac{g(x_0)-g(x)}{g(x_0)g(x)}=-\frac{(x-x_0)\Delta_{g,x_0}(x)}{g(x_0)g(x)}\] and thus \[\lim_{x\to x_0}\frac1{x-x_0}\left(\frac1{g(x)}-\frac1{g(x_0)}\right)=-\frac{g'(x_0)}{g^2(x_0)}.\]
\(\Box\)
Example 2.
For a constant \(c\) and a differentiable function \(f:I\to\mathbb{R}\) we have \[(cf)'(x)=c'f(x)+cf'(x)=cf'(x).\]
Let \(n\in\mathbb{N}\) and \(f:\mathbb{R}\to\mathbb{R}, x\mapsto x^n\). Then \(f'(x)=nx^{n-1}\).
Induction basis: \(n=1\). \(f'(x)=(1\cdot x)'\underset{a)}{=}1=1\cdot x^{1-1}\).
Induction step: \(n>1\): Using the product rule we immediately get \[f'(x)=(x^n)'=(x\cdot x^{n-1})' = x'(x^{n-1})+x(x^{n-1})'\underset{\text{Ind. Hyp.}}{=} x^{n-1}+x\cdot (n-1)x^{n-2}=nx^{n-1}\]\(f(x)=x^{-n}=\frac1{x^n}\), \(n\in\mathbb{N}\). Then for \(x\in\mathbb{R}\backslash\{0\}\) \[f'(x)=\frac{-nx^{n-1}}{x^{2n}}=-n\frac{1}{x^{n+1}}=-nx^{-n-1}\] is true.
\(f(x)=x\cdot \exp(x)\). Then for \(x\in\mathbb{R}\) \[f'(x)=1\cdot\exp(x)+x\cdot\exp(x)=(1+x)\exp(x).\]
For a real polynomial \(f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0\) holds \[f'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\ldots+2a_2x+a_1.\]
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