Question

срочно даю 50 балів
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Answer 1

Решение.

Все производные табличные, которые надо знать наизусть .

$\bf y=-5x^3-4x-5\ \ ,\ \ \ y'=-15x^2-4\\\\y=-sinx-2\, tgx-cosx\ \ ,\ \ y'=-cosx-\dfrac{2}{cos^2x}+sinx\\\\y=-3x^2-2\, sinx\ \ ,\ \ \ y'=-6x-2\, cosx\\\\y=2x^4-3x^3+2x^5-sinx\ \ ,\ \ y'=8x^3-9x^2+10x^4-cosx\\\\y=4x^7-7x^{-2}-4x\ \ ,\ \ \ y'=28x^3+14x^{-3}-4\\\\y=-3\sqrt{x}+4x^4\ \ ,\ \ \ y'=-\dfrac{3}{2\sqrt{x}}+16x^3\\\\y=-sinx-cosx-ctgx\ \ ,\ \ y'=-cosx+sinx+\dfrac{1}{sin^2x}$

$\bf y=\sqrt[4]{x^3}+\sqrt[5]{x^2}=x^{\frac{3}{4}}+x^{\frac{2}{5}}\ \ ,\ \ y'=\dfrac{3}{4}\, x^{-\frac{1}{4}}+\dfrac{2}{5}\, x^{-\frac{3}{5}}=\dfrac{3}{4\sqrt[4]{x}}+\dfrac{2}{5\sqrt[5]{x^3}}$

$\bf y=\sqrt[3]{x^{-2}}+\dfrac{2}{x}-5x\ \ ,\ \ y'=-\dfrac{2}{3}\, x^{-\frac{5}{3}}-\dfrac{2}{x^2}-5=-\dfrac{2}{3\sqrt[3]{x^5}}-\dfrac{2}{x^2}-5\\\\\\y=\dfrac{4x}{-5x^2+3} \ \ ,\ \ y'=\dfrac{4(-5x^2+3)-4x\cdot (-10x)}{(3-5x^2)^2}=\dfrac{20x^2+12}{(3-5x^2)^2}\\\\\\y=\dfrac{4x^4-2x}{7x}=\dfrac{4x^3-2}{7}=\dfrac{1}{7}\cdot (4x^3-2)\ \ ,\ \ \ y'=\dfrac{1}{7}\cdot 12x^2=\dfrac{12}{7}\, x^2$