Question

Помогите пожалуйста решить производные срочно!!

Answer 1

$1)f(x)=\sqrt{x\sqrt{x\sqrt{x}}}=\sqrt{x\sqrt{x*x^{\frac{1}{2}}}}=\sqrt{x\sqrt{x^{\frac{3}{2}}}}=\sqrt{x*x^{\frac{3}{4}}}=\sqrt{x^{\frac{7}{4}}}\\\\f'(x)=(\sqrt{x^{\frac{7}{4}}})'=\frac{1}{2\sqrt{x^{\frac{7}{4}}}}*(x^{\frac{7}{4}})'=\frac{\frac{7}{4}x^{\frac{3}{4}}}{2x^{\frac{7}{8}}}=\frac{7}{8} x^{\frac{3}{4}-\frac{7}{8}}=\frac{7}{8}x^{-\frac{1}{8}}=\frac{7}{8\sqrt[8]{x}}$

$2)f(x)=5x^{4}-7x^{3}-2\sqrt{x} +\frac{5}{x}\\\\f'(x)=5(x^{4})'-7(x^{3})'-2*(\sqrt{x})'+5*(\frac{1}{x})'=5*4x^{3}-7*3x^{2}-2*\frac{1}{2\sqrt{x}}+5*(-\frac{1}{x^{2}})=20x^{3}-21x^{2}-\frac{1}{\sqrt{x}}-\frac{5}{x^{2}}$

$3)f(x)=(2x^{3} +3x)^{4}\\\\f'(x)=4(2x^{3}+3x)^{3}*(2x^{3}+3x)'=4(2x^{3}+3x)^{3}*(6x^{2}+3)$

$4)f(x)=3x^{2}\sqrt{x}+4x^{7}-\frac{8}{x^{4}}+3x^{3}=3x^{2}*x^{\frac{1}{2}}+4x^{7}-8x^{-4}+3x^{3}=3x^{\frac{5}{2}}+4x^{7}-8x^{-4} +3x^{3}\\\\f'(x)=3(x^{\frac{5}{2}})'+4(x^{7})'-8(x^{-4})'+3(x^{3})'=3*\frac{5}{2}x^{\frac{3}{2}}+4*7x^{6}-8*(-4x^{-5})+3*3x^{2}=7,5x\sqrt{x}+28x^{6} +\frac{32}{x^{5}}+9x^{2}$

$5)f(x)=\frac{2x+3}{x^{2}+4}\\\\f'(x)=\frac{(2x+3)'*(x^{2}+4)-(2x+3)*(x^{2}+4)'}{(x^{2}+4)^{2}} =\frac{2*(x^{2}+4)-2x*(2x+3) }{(x^{2}+4)^{2}}=\frac{2x^{2}+8-4x^{2}-6x}{(x^{2}+4)^{2}}=\frac{-2x^{2}-6x+8 }{(x^{2}+4)^{2}}$

$6)f(x)=6x^{3}-\frac{4}{x^{4}}+\frac{2}{x^{6}}-\sqrt{3}=6x^{3}-4x^{-4}+2x^{-6}-\sqrt{3}\\\\f'(x)=6(x^{3})'-4(x^{-4})' +2(x^{-6})'-(\sqrt{3})'=6*3x^{2}-4*(-4x^{-5} )+2*(-6x^{-7} )=18x^{2}+\frac{16}{x^{5}}-\frac{12}{x^{7}}$

$7)f(x)=x^{2}*\sqrt{1+x}\\\\f'(x)=(x^{2})'*\sqrt{1+x}+x^{2}*(\sqrt{1+x})'=2x\sqrt{1+x}+x^{2}*\frac{1}{2\sqrt{1+x}}=\frac{4x(1+x)+x^{2}}{2\sqrt{1+x}} =\frac{4x+4x^{2}+x^{2}}{2\sqrt{1+x}}=\frac{5x^{2}+4x }{2\sqrt{1+x} }$

Answer 2

Ответ: во вложении Объяснение: