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July 2022 1 4 Report

(PLEASE) найдите производную функций

Пожалуйста, с решение, тк пытаюсь разобрать новую тему [tex]f(x) = 2x^5-\frac{x^3}{3} +3x^2-4\\\\f(x) = (3x-5)\sqrt{x} \\\\f(x) = \frac{x^2+9x}{x-4} \\\\f(x) = \frac{2}{x^3} -\frac{3}{x^6}[/tex]

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Universalka

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$1)f(x)=2x^{5}-\frac{x^{3} }{3} +3x^{2}-4\\\\f'(x)=2(x^{5})'-\frac{1}{3}(x^{3})'+3(x^{2})'-4'=2*5x^{4}-\frac{1}{3}*3x^{2}+3*2x-0=10x^{4}-x^{2}+6x\\\\2)f(x)=(3x-5)\sqrt{x}\\\\f'(x)=(3x-5)'*\sqrt{x}+(3x-5)*(\sqrt{x})'=3\sqrt{x}+(3x-5)*\frac{1}{2\sqrt{x} }=3\sqrt{x} +\frac{3x-5}{2\sqrt{x} }=\frac{6x+3x-5}{2\sqrt{x} }=\frac{9x-5}{2\sqrt{x} }$

$3)f(x)=\frac{x^{2} +9x}{x-4}\\\\f'(x)=\frac{(x^{2}+9x)'*(x-4)-(x^{2}+9x)*(x-4)'}{(x-4)^{2} } =\frac{(2x+9)(x-4)-(x^{2}+9x) }{(x-4)^{2} }=\frac{2x^{2}-8x+9x-36-x^{2}-9x}{(x-4)^{2} }=\frac{x^{2}-8x-36 }{(x-4)^{2} }$

$4)\frac{2}{x^{3} }-\frac{3}{x^{6} }=\frac{2x^{3}-3 }{x^{6} }\\\\f'(x)=\frac{(2x^{3}-3)'*x^{6}-(2x^{3}-3)*(x^{6})'}{(x^{6})^{2}}=\frac{6x^{2}*x^{6}-6x^{5}(2x^{3}-3)}{x^{12} }=\frac{6x^{8}-12x^{8}+18x^{5}}{x^{12} }=\frac{18x^{5}-6x^{8}}{x^{12} }=\frac{x^{5}(18-6x^{3})}{x^{12} }=\frac{18-x^{3}}{x^{7} }$

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