Знайти похідні таких функцій: 1) y = tg^2((2x + 7)^4) 2) y = (9x-3)/(x^2 + 3x - 1). Created by SlavaShev. algebra-ru

(x^2 + 3x - 1)...

SlavaShev @SlavaShev

July 2022 1 11 Report

Знайти похідні таких функцій:

1) y = tg^2((2x + 7)^4)
2) y = (9x-3)/(x^2 + 3x - 1)

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$y=tg^{2}((2x+7) ^{4})\\\\y'=2tg((2x+7)^{4})*(tg(2x+7) ^{4})'=2tg((2x+7) ^{4})* \frac{1}{Cos^{2}(2x+7)^{4}} *((2x+7) ^{4})'= \frac{2Sin(2x+7)^{4}}{Cos^{3}(2x+7)^{4}}*4(2x+7) ^{3}*(2x+7)'=\frac{16Sin(2x+7)^{4}*(2x+7)^{3}}{Cos^{3}(2x+7)^{4}}$

$y=\frac{9x-3}{x^{2}+3x-1}\\\\y'=\frac{(9x-3)'*(x^{2}+3x-1)-(9x-3)*(x^{2}+3x-1)'}{(x^{2}+3x-1)^{2}}= \frac{9*(x^{2}+3x-1)-(9x-3)*(2x+3)}{(x^{2}+3x-1)^{2}}=$ $=\frac{9x^{2}+27x-9-18x^{2}-27x+6x+9}{(x^{2}+3x-1)^{2}}= \frac{-9x^{2}+6x} {(x^{2}+3x-1)^{2}}=- \frac{3x(3x-2)}{(x^{2}+3x-1)^{2}}$

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9 x 2 xxdx

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