Question

Найти производную
$x {}^{3} + \frac{1}{x} - 1$
$16 \sqrt{x} - 4x {}^{2}$
$e {}^{x} + \sin(x)$
$e {}^{5} - 3x$

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Answer 1

$y=x^3+\dfrac{1}{x} -1$

$y'=3x^2-\dfrac{1}{x^2}-0=3x^2-\dfrac{1}{x^2}$

$y=16\sqrt{x} -4x^2$

$y'=16\cdot\dfrac{1}{2\sqrt{x} } -4\cdot2x=\dfrac{8}{\sqrt{x} } -8x$

$y=e^x+\sin x$

$y'=e^x+\cos x$

$y=e^5-3x$

$y'=0-3=-3$