Question

Найдите производную функции
$y=\frac{x^9}{9}+ \sqrt{2cosx}+tg \frac{\pi} {4}-2x^4$

Answer 1

$y = \frac{x^9}9 + \sqrt{2cosx} + tg\frac\pi4 - 2x^4\\y' = (\frac{x^9}9)' + (\sqrt{2cosx})' + (tg\frac\pi4)' - (2x^4)'\\(x^n)' = nx^{n-1}*x'; (\sqrt x)' = \frac{x'}{2\sqrt x}; (tgx)' = \frac{x'}{cos^2x}\\y' = x^8 + \frac{\sqrt2}{2\sqrt{cosx}} + \frac1{cos^2\frac\pi4} - 8x^3$

Answer 2

$y=\frac{x^9}{9}+\sqrt{2}cos^{0,5}x+tg \frac{\pi} {4}-2x^4;\\y'=\frac{9}{9}x^8-\frac{\sqrt{2}sinx}{2cos^{0,5}x}+0-2*4x^3=\\=x^8-\frac{sinx}{\sqrt{2cosx}}-8x^3$