Question

145
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5 6 7 8​

Answer 1

Объяснение:

$\displaystyle 5)y=arctg(3x-4)\\\\y'=\frac{(3x-4)'}{1+(3x-4)^2} =\frac{3}{1+(3x-4)^2}$

$\displaystyle 6)y=sin(arctg\;2x)\\\\y'=cos(arctg\;2x)*(arctg\;2x)'=cos(arctg\;2x)*\frac{(2x)'}{1+4x^2} =\\\\=\frac{2cos(arctg\;2x)}{1+4x^2}$

$\displaystyle 7)y=sin^3x+cos^3x\\\\y'=3sin^2x*(sinx)'+3cos^2x*(cosx)'=\\\\=3sin^2x\;cosx+3cos^2x(-sin\;x)=\\\\=3sin^2x\;cosx-3cos^2x\;sinx=3sinx\;cosx(sinx-cosx)=\\\\=\frac{3}{2}sin2x(sinx-cosx)$

$\displaystyle 8)y=e^{sin(cosx)}\\\\y'=e^{sin(cosx)}*(sin(cosx))'=e^{sin(cosx)}cos(cosx)*(cosx)'=\\\\=e^{sin(cosx)}cos(cosx)(-sinx)=-e^{sin(cosx)}cos(cosx)\;sinx$

Использованы формулы:

$\displaystyle (c)'=0,\;\;\;c-const\\(cx)'=c\\\\(argtg\;x)'=\frac{1}{1+x^2}\\\\(sin\;x)'=cos\;x\\(cos\;x)'=-sin\;x$