Question

A) y=x^2-3x²+2tgxb) y=2x^10+3cosx -√x Найдите производную

Answer 1

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$\displaystyle (tgx)'= \frac{sinx'}{cosx'}= \frac{sinx'cosx-sinxcosx'}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}= \frac{1}{cos^2x}$
$\displaystyle (\sqrt x)'= (x^{1/2})'= \frac{1}{2}x^{1/2-1}= \frac{1}{2}x^{- \frac{1}{2}}=\frac{1}{2 \sqrt x}$
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$\boxed{a.} y=x^2-3x^2+2tgx$
$y=-2x^2+2tgx$
$\displaystyle y'=-4x+(2'tgx+2tgx')=-4x+ \frac{2}{cos^2x}$
$\boxed{b.} y=2x^1^0+3cosx- \sqrt x$
$\displaystyle y'=20x^9-3sinx- \frac{1}{2 \sqrt x}$