Question

Помогите с производными

Answer 1

$\displaystyle\bf\\1)\\\\f(x)=\frac{3}{\sqrt[3]{x} } -6\sqrt[3]{x^{4} } =\frac{3}{x^{\frac{1}{3} } } -6x^{\frac{4}{3} } =3x^{-\frac{1}{3} } -6x^{\frac{4}{3} } \\\\\\f'(x)=3\cdot\Big(-\frac{1}{3} \Big)\cdot x^{-\frac{4}{3} } -6\cdot\Big(\frac{4}{3} \Big)\cdot x^{\frac{1}{3} } =-\frac{1}{x^{\frac{4}{3} } } -8x^{\frac{1}{3} } =-\frac{1}{\sqrt[3]{x^{4} } } -8\sqrt[3]{x} \\\\\\2)\\\\f(x)=e^{3x+2} \\\\f'(x)=e^{3x+2} \cdot (3x+2)'=3e^{3x+2} \\\\\\3)\\\\f(x)=x\sqrt{x^{2} -3x+4}$

$\displaystyle\bf\\f'(x)=x'\cdot\sqrt{x^{2} -3x+4} +x\cdot(\sqrt{x^{2} -3x+4} )'=\\\\\\=1\cdot\sqrt{x^{2} -3x+4} +x\cdot\frac{1}{2\sqrt{x^{2} -3x+4} } \cdot(x^{2} -3x+4)'=\\\\\\=\sqrt{x^{2} -3x+4} +\frac{x}{2\sqrt{x^{2} -3x+4} } \cdot(2x-3)=\\\\\\=\sqrt{x^{2} -3x+4} +\frac{2x^{2} -3x}{2\sqrt{x^{2} -3x+4} }=\frac{2(x^{2} -3x+4)+2x^{2} -3x}{2\sqrt{x^{2} -3x+4} } =\\\\\\=\frac{2x^{2} -6x+8+2x^{2} -3x}{2\sqrt{x^{2} -3x+4} } =\frac{4x^{2} -9x+8}{2\sqrt{x^{2} -3x+4} } =$

$\displaystyle\bf\\=\frac{2x^{2} -4,5x+4}{\sqrt{x^{2} -3x+4} }$