от ammornil » 17 Авг 2024, 11:22
[tex]f(x)=\underbrace{\left[ \frac{\left((\sqrt{3}+x)(\sqrt[3]{3}-\sqrt[3]{x^{2}})^{-1}-(\sqrt[3]{3x}-\sqrt[3]{x^{2}\sqrt{3}})(\sqrt[3]{x}-\sqrt[6]{3})^{-2} \right)}{\left((\sqrt[12]{3}-\sqrt[6]{x})(\sqrt[3]{x}-\sqrt[6]{x\sqrt{3}}-2\sqrt[6]{3}) \right)} \right]^{-1}}_{g(x)}+\underbrace{2\sqrt{3}\log_{3}{\sqrt[6]{3}}}_{h(x)} \\ f(x)=g(x)+h(x) \Rightarrow f'(x)=g'(x)+h'(x) \\ h'(x)=0 \Rightarrow f'(x)=g'(x)[/tex]
Нека направим някои преобразувания извън основното условие, за да си спестим преписвания.
[tex](\sqrt{3}+x)(\sqrt[3]{3}-\sqrt[3]{x^{2}})^{-1}=\frac{\sqrt{3}+x}{\sqrt[3]{3}-\sqrt[3]{x^{2}}}=\frac{\left(3^{\frac{1}{6}}\right)^{3}+\left(x^{\frac{1}{3}}\right)^{3}}{\left(3^{\frac{1}{6}}\right)^{2}-\left(x^{\frac{1}{3}}\right)^{2}} \\ \frac{\sqrt[3]{3x}-\sqrt[3]{x^{2}\sqrt{3}}}{(\sqrt[3]{x}-\sqrt[6]{3})^{2}}=\frac{3^{\frac{1}{3}}x^{\frac{1}{3}}-(x^{\frac{1}{3}})^{2}3^{\frac{1}{6}}}{(x^{\frac{1}{3}})^{2}-2x^{\frac{1}{3}}3^{\frac{1}{6}}+3^{\frac{1}{3}}} \\ (\sqrt[12]{3}-\sqrt[6]{x})(\sqrt[3]{x}-\sqrt[6]{x\sqrt{3}}-2\sqrt[6]{3})=(3^{\frac{1}{12}}-x^{\frac{1}{6}})[(x^{\frac{1}{6}})^{2}-x^{\frac{1}{6}}3^{\frac{1}{12}}-2\cdot{}3^{\frac{1}{6}}][/tex]
Забелязвам, че е възможно [tex]3^{\frac{1}{6}}=u[/tex] и [tex]x^{\frac{1}{3}}=v[/tex] да са удобни за субституция и пробвам.
[tex]g(x)=\left[ \frac{\left((\sqrt{3}+x)(\sqrt[3]{3}-\sqrt[3]{x^{2}})^{-1}-(\sqrt[3]{3x}-\sqrt[3]{x^{2}\sqrt{3}})(\sqrt[3]{x}-\sqrt[6]{3})^{-2} \right)}{\left((\sqrt[12]{3}-\sqrt[6]{x})(\sqrt[3]{x}-\sqrt[6]{x\sqrt{3}}-2\sqrt[6]{3}) \right)} \right]^{-1} \\ g(x) = \frac{\left((\sqrt[12]{3}-\sqrt[6]{x})(\sqrt[3]{x}-\sqrt[6]{x\sqrt{3}}-2\sqrt[6]{3}) \right)}{\left((\sqrt{3}+x)(\sqrt[3]{3}-\sqrt[3]{x^{2}})^{-1}-(\sqrt[3]{3x}-\sqrt[3]{x^{2}\sqrt{3}})(\sqrt[3]{x}-\sqrt[6]{3})^{-2} \right)} \\ g(x)= \frac{(3^{\frac{1}{12}}-x^{\frac{1}{6}})[(x^{\frac{1}{6}})^{2}-x^{\frac{1}{6}}3^{\frac{1}{12}}-2\cdot{}3^{\frac{1}{6}}]}{\frac{\left(3^{\frac{1}{6}}\right)^{3}+\left(x^{\frac{1}{3}}\right)^{3}}{\left(3^{\frac{1}{6}}\right)^{2}-\left(x^{\frac{1}{3}}\right)^{2}}-\frac{3^{\frac{1}{3}}x^{\frac{1}{3}}-(x^{\frac{1}{3}})^{2}3^{\frac{1}{6}}}{(x^{\frac{1}{3}})^{2}-2x^{\frac{1}{3}}3^{\frac{1}{6}}+3^{\frac{1}{3}}}} \\ g(x)= \frac{(u^{\frac{1}{2}}-v^{\frac{1}{2}})[(v^{\frac{1}{2}})^{2}-v^{\frac{1}{2}}u^{\frac{1}{2}}-2\cdot{}u]}{\frac{\left(u\right)^{3}+\left(v\right)^{3}}{\left(u\right)^{2}-\left(v\right)^{2}}-\frac{u^{2}v-v^{2}u}{v^{2}-2vu+u^{2}}} \\ g(x)=\frac{(u^{\frac{1}{2}}-v^{\frac{1}{2}})[(v^{\frac{1}{2}})^{2}-v^{\frac{1}{2}}u^{\frac{1}{2}}-2\cdot{}u]}{\frac{(u+v)(u^{2}-uv+v^{2})}{(u-v)(u+v)}-\frac{uv(u-v)}{(u-v)^{2}}} \\ g(x)=\frac{(u^{\frac{1}{2}}-v^{\frac{1}{2}})[(v^{\frac{1}{2}})^{2}-v^{\frac{1}{2}}u^{\frac{1}{2}}-2\cdot{}u]}{\frac{u^{2}-uv+v^{2}}{u-v}-\frac{uv}{u-v}} \\ g(x)=\frac{(u^{\frac{1}{2}}-v^{\frac{1}{2}})[(v^{\frac{1}{2}})^{2}-v^{\frac{1}{2}}u^{\frac{1}{2}}-2\cdot{}u]}{\frac{u^{2}-uv+v^{2}-uv}{u-v}} \\ g(x)=\frac{(u^{\frac{1}{2}}-v^{\frac{1}{2}})[(v^{\frac{1}{2}})^{2}-v^{\frac{1}{2}}u^{\frac{1}{2}}-2\cdot{}u]}{\frac{(u-v)^{2}}{u-v}} \\ g(x)=\frac{(u^{\frac{1}{2}}-v^{\frac{1}{2}})[(v^{\frac{1}{2}})^{2}-v^{\frac{1}{2}}u^{\frac{1}{2}}-2\cdot{}u]}{(u^{\frac{1}{2}})^{2}-(v^{\frac{1}{2}})^{2}} \\ g(x)=\frac{(u^{\frac{1}{2}}-v^{\frac{1}{2}})[(v^{\frac{1}{2}})^{2}-v^{\frac{1}{2}}u^{\frac{1}{2}}-2\cdot{}u]}{(u^{\frac{1}{2}}-v^{\frac{1}{2}})(u^{\frac{1}{2}}+v^{\frac{1}{2}})} \\ g(x)=\frac{v-v^{\frac{1}{2}}u^{\frac{1}{2}}-2\cdot{}u}{u^{\frac{1}{2}}+v^{\frac{1}{2}}}=\frac{x^{\frac{1}{3}}-x^{\frac{1}{6}}3^{\frac{1}{12}}-2\cdot{}3^{\frac{1}{6}}}{3^{\frac{1}{12}}+x^{\frac{1}{6}}} \\[/tex]$$ g(x)=\frac{\sqrt[3]{x}-\sqrt[12]{3}\sqrt[6]{x}-2\sqrt[6]{3}}{\sqrt[12]{3}+\sqrt[6]{x}} $$
[tex]h(x)=2\sqrt{3}\log_{3}{\sqrt[6]{3}}=2\sqrt{3}\cdot{\frac{1}{6}}=\frac{\sqrt{3}}{3}[/tex]
$$ f(x)=\frac{\sqrt[3]{x}-\sqrt[12]{3}\sqrt[6]{x}-2\sqrt[6]{3}}{\sqrt[12]{3}+\sqrt[6]{x}}+\frac{\sqrt{3}}{3} $$
Можете ли да намерите производната на тази функция?
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[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]