Question

Вычислите производные функций

Answer 1

Ответ:

1.

$y = {x}^{2} \cos(2x)$

$y' = ( {x}^{2} )' \cos(2x) + ( \cos(2x)) ' \times {x}^{2} = \\ = 2x \cos(2x) - 2 \sin(2x) \times {x}^{2} = \\ = 2x( \cos(2x) - x \sin(2x))$

2.

$y = {2}^{tg( \frac{1}{x}) }$

$y' = ln(2) \times {2}^{tg \frac{1}{x} } \times \frac{1}{ { \cos }^{2} ( \frac{1}{x}) } \times ( - {x}^{ - 2} ) = \\ = - \frac{ ln(2) \times {2}^{tg \frac{1}{x} } }{ {x}^{2} { \cos}^{2} ( \frac{1}{x}) }$

3.

$y = {e}^{x} (1 + ctg \frac{x}{2} )$

$y' = {e}^{x} (1 + ctg \frac{x}{2} ) + {e}^{x} \times ( - \frac{1}{ \sin ^{2} ( \frac{x}{2} ) } ) \times \frac{1}{2} = \\ = {e}^{x} (1 + ctg \frac{x}{2} - \frac{1}{2 { \sin }^{2} \frac{x}{2} } )$

4.

$y = {ln}^{3} ( {x}^{2} )$

$y' = 3 {ln}^{2} ( {x}^{2} ) \times \frac{1}{ {x}^{2} } \times 2x = \\ = \frac{6ln ^{2}( {x}^{2} ) }{x}$

5.

$y = ( { ln(2x)) }^{ \sqrt{x} }$

$y' = ( ln(y)) ' \times y$

$( ln(y))' = ( ln( ln( {2x)}^{ \sqrt{x} } ) ) )' = \\ = ( \sqrt{x} \times ln( ln(2x) ) ) '= \\ = \frac{1}{2 \sqrt{x} } \times ln( ln(2x) ) + \frac{1}{ ln(2x) } \times \frac{1}{2x} \times 2 \times \sqrt{x} = \\ = \frac{ ln( ln(2x) ) }{ \sqrt{x} } + \frac{1}{ \sqrt{x} ln(2x) }$

$y' = {( ln(2x)) }^{ \sqrt{x} } \times ( \frac{ ln( ln(2x) ) }{2 \sqrt{x} } + \frac{1}{ \sqrt{x} ln(2x) } )$

Answer 2

$1)\ \ y=x^2\cdot cos2x\ \ \ ,\ \ \ \ \ (uv)'=u'v+uv'\\\\y'=2x\cdot cos2x-2x^2sin2x=2x\cdot (cos2x-x\cdot sin2x)\\\\\\2)\ \ y=e^{x}\cdot (1+ctg\dfrac{x}{2})\\\\y'=e^{x}\cdot (1+ctg\dfrac{x}{2})+e^{x}\cdot \dfrac{-1}{sin^2\frac{x}{2}}\cdot \dfrac{1}{2}=e^{x}\cdot \Big(1+ctg\dfrac{x}{2}-\dfrac{1}{2sin^2\frac{x}{2}}\Big)$

$3)\ \ y=2^{tg\frac{1}{x}}\\\\y'=2^{tg\frac{1}{x}}\cdot ln2\cdot \dfrac{1}{cos^2\frac{1}{x}}\cdot \dfrac{-1}{x^2}=-\dfrac{2^{tg\frac{1}{x}}\cdot ln2}{x^2\cdot cos^2\frac{1}{x}}$

$4)\ \ y=ln^3(x^2)\ \ ,\ \ \ \ \ \ \ \ (u^3)'=3u^2\cdot u'\ ,\ \ u=ln(x^2)\\\\y'=3\cdot ln^2(x^2)\cdot (ln(x^2))'=3\cdot ln^2(x^2)\cdot \dfrac{2x}{x^2}=\dfrac{6\cdot ln^2(x^2)}{x}$

$5)\ \ y=(ln2x)^{\sqrt{x}}\ \ \ \to \ \ \ \ ln\, y=ln\Big((ln2x)^{\sqrt{x}}\Big)\\\\ln\, y=\sqrt{x} \cdot ln(ln2x)\\\\\Big(ln\, y\Big)'=\Big(\sqrt{x} \cdot ln(ln2x)\Big)'\\\\\dfrac{y'}{y}=\dfrac{1}{2\sqrt{x}}\cdot ln(ln2x)+\sqrt{x}\cdot \dfrac{1}{ln2x}\cdot \dfrac{1}{2x}\cdot 2\\\\\dfrac{y'}{y}=y\cdot \Big(\dfrac{1}{2\sqrt{x}}\cdot ln(ln2x)+\sqrt{x}\cdot \dfrac{1}{ln2x}\cdot \dfrac{1}{x}\Big)\\\\y'=(ln2x)^{\sqrt{x}}\cdot \Big(\dfrac{1}{2\sqrt{x}}\cdot ln(ln2x)+\dfrac{1}{\sqrt{x}\cdot ln2x}\Big)$