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Constant Multiple Rule

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The Constant Multiple Rule. If $$${c}$$$ is a constant anf $$${f{}}$$$ is a differentiable function then $$${\left({c}{f{{\left({x}\right)}}}\right)}'={c}{\left({f{{\left({x}\right)}}}\right)}'$$$.

Proof. By definition $$${\left({c}{f{{\left({x}\right)}}}\right)}'=\lim_{{{h}\to{0}}}\frac{{{c}{f{{\left({x}+{h}\right)}}}-{c}{f{{\left({x}\right)}}}}}{{h}}={c}\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}={c}{f{'}}{\left({x}\right)}$$$.

Example 1. Find $$${f{'}}{\left({x}\right)}$$$ if $$${f{{\left({x}\right)}}}={2}\cdot{{3}}^{{x}}$$$.

$$${f{'}}{\left({x}\right)}={\left({2}\cdot{{3}}^{{x}}\right)}'={2}{\left({{3}}^{{x}}\right)}'={2}\cdot{{3}}^{{x}}{\ln{{\left({3}\right)}}}$$$.

Example 2. Find $$${f{'}}{\left({x}\right)}$$$ if $$${f{{\left({x}\right)}}}=\frac{{1}}{{{2}\sqrt{{{x}}}}}$$$.

$$${f{'}}{\left({x}\right)}={\left(\frac{{1}}{{{2}\sqrt{{{x}}}}}\right)}'=\frac{{1}}{{2}}{\left(\frac{{1}}{\sqrt{{{x}}}}\right)}'=\frac{{1}}{{2}}{\left(\frac{{1}}{{{{x}}^{{\frac{{1}}{{2}}}}}}\right)}'=\frac{{1}}{{2}}{\left({{x}}^{{-\frac{{1}}{{2}}}}\right)}'=\frac{{1}}{{2}}\cdot{\left(-\frac{{1}}{{2}}\right)}{{x}}^{{-\frac{{1}}{{2}}-{1}}}=-\frac{{1}}{{4}}{{x}}^{{-\frac{{3}}{{2}}}}$$$.