Question

Вычислите производную
братцы ✊​

Answer 1

$3x^{2}+\frac{1}{x^{2}};$  $\frac{2}{\sqrt{x}};$  $32x+2;$  $-sinx;$

Объяснение:

$2) (x^{3}-\frac{1}{x})';$

$(a+b)'=a'+b';\\(a-b)'=a'-b';$

$(x^{3})'-(\frac{1}{x})';$

$(x^{\alpha})'=\alpha x^{\alpha -1};$

$3x^{3-1}-(\frac{1}{x})'=3x^{2}-(\frac{1}{x^{1}})';$

$\frac{1}{x^{n}}=x^{-n};$

$3x^{2}-(x^{-1})'=3x^{2}-(-1)x^{-1-1}=3x^{2}+1x^{-2}=3x^{2}+x^{-2}=3x^{2}+\frac{1}{x^{2}};$

$4) (4\sqrt{x})';$

$(Cx)'=C(x)', C-const;$

$(4\sqrt{x})'=4(\sqrt{x})'=4(\sqrt[2]{x^{1}})';$

$\sqrt[n]{x^{m}}=x^{\frac{m}{n}};$

$4(\sqrt[2]{x^{1}})'=4(x^{\frac{1}{2}})'=4*\frac{1}{2}*x^{\frac{1}{2}-1}=2x^{-\frac{1}{2}}=2(x^{-1})^{\frac{1}{2}}=2(\frac{1}{x})^{\frac{1}{2}}=2\sqrt[2]{(\frac{1}{x})^{1}}=$

$=2\sqrt{\frac{1}{x}}=2\frac{\sqrt{1}}{\sqrt{x}}=2\frac{1}{\sqrt{x}}=\frac{2*1}{\sqrt{x}}=\frac{2}{\sqrt{x}};$

$6) ((2x-1)*(8x+5))'=(2x*8x+2x*5-1*8x-1*5)'=(16x^{2}+10x-8x-$

$-5)'=(16x^{2}+2x-5)'=(16x^{2})'+(2x)'-5';$

$C'=0, C-const;$

$(16x^{2})'+(2x)'-5'=16*(x^{2})'+2x'-0=16*2*x^{2-1}+2*1=32x^{1}+2=$

$=32x+2;$

$8) (\frac{sinx}{tgx})'=(\frac{sinx}{\frac{sinx}{cosx}})'=(\frac{sinx*cosx}{sinx})'=(cosx)'=-sinx;$