Question

Đã trả lời bởi chuyên gia

Answer 1

a) $y^{'} = {[{(x^{2} - \frac{2}{x} + 4 \sqrt{x})}^{3}]}^{'}$

$= 3 {(x^{2} - \frac{2}{x} + 4 \sqrt{x})}^{2} {(x^{2} - \frac{2}{x} + 4 \sqrt{x})}^{'}$

$= 3 {(x^{2} - \frac{2}{x} + 4 \sqrt{x})}^{2} (2 x + \frac{2}{x^{2}} + \frac{2}{\sqrt{x}})$

$= 6 {(x^{2} - \frac{2}{x} + 4 \sqrt{x})}^{2} (x + \frac{1}{x^{2}} + \frac{1}{\sqrt{x}})$

Vậy $y^{'} = 6 {(x^{2} - \frac{2}{x} + 4 \sqrt{x})}^{2} (x + \frac{1}{x^{2}} + \frac{1}{\sqrt{x}})$

b) y' = [2^x + log₃(1 – 2x)]' = $2^{x} \ln 2 + \frac{{(1 - 2 x)}^{'}}{(1 - 2 x) \ln 3}$ $= 2^{x} \ln 2 - \frac{2}{(1 - 2 x) \ln 3}$

Vậy $y^{'} = 2^{x} \ln 2 - \frac{2}{(1 - 2 x) \ln 3}$

c) $y^{'} = {(\frac{1 - 2 x}{x^{2} + 1})}^{'}$ $= \frac{{(1 - 2 x)}^{'} (x^{2} + 1) - (1 - 2 x) {(x^{2} + 1)}^{'}}{{(x^{2} + 1)}^{2}}$

$= \frac{- 2 (x^{2} + 1) - 2 x (1 - 2 x)}{{(x^{2} + 1)}^{2}}$ $= \frac{- 2 x^{2} - 2 - 2 x + 4 x^{2}}{{(x^{2} + 1)}^{2}}$ $= \frac{2 x^{2} - 2 x - 2}{{(x^{2} + 1)}^{2}}$

Vậy $y^{'} = \frac{2 x^{2} - 2 x - 2}{{(x^{2} + 1)}^{2}}$

d) y' = (sin2x + cos²3x)' = cos2x×(2x)' + 2cos3x×(cos3x)'

= 2cos2x – 6cos3xsin3x = 2cos2x – 3sin6x.

Vậy y' = 2cos2x – 3sin6x.

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Answer 2