Verified answer

Ответ:

Производные функций:

1) [tex]f'(x) = \dfrac{1}{x^{2} -1}[/tex]

2) [tex]f'(x) = 21 \cos 7x \sin^{2} 7x[/tex]

3) [tex]f'(x) = -16\ \text{tg} \ 8x[/tex]

Примечание:

По таблице производных:

[tex]\boxed{(\ln x)' = \frac{1}{x} }[/tex]

[tex]\boxed{(\sqrt{x} )' = \frac{1}{2\sqrt{x} } }[/tex]

[tex]\boxed{C' = 0}[/tex], где [tex]C \in \mathbb R[/tex]

[tex]\boxed{(x^{n})' = nx^{n - 1}}[/tex]

[tex]\boxed{(\sin x)' = \cos x}[/tex]

[tex]\boxed{(\cos x)' = -\sin x}[/tex]

Правила дифференцирования:

[tex](f \pm g)' = f' \pm g'[/tex]

[tex](fg)' = f'g + fg'[/tex]

[tex]\bigg(\dfrac{f}{g} \bigg)' = \dfrac{f'g - fg'}{g^{2}}[/tex]

[tex]f(g) = g'f'(g)[/tex]

[tex](kf)' = k(f')[/tex], где [tex]k \in \mathbb R[/tex]

[tex]f,g \ -[/tex] функции одной переменной

Пошаговое объяснение:

1) [tex]f(x) = \ln \sqrt{\dfrac{x - 1}{x + 1} }[/tex]

[tex]f'(x) = \bigg( \ln \sqrt{\dfrac{x - 1}{x + 1} } \bigg)' = \dfrac{\Bigg(\sqrt{\dfrac{x - 1}{x + 1} } \Bigg)'}{\sqrt{\dfrac{x - 1}{x + 1} } } =\dfrac{\bigg( \dfrac{x - 1}{x + 1}\bigg)'}{2\sqrt{\dfrac{x - 1}{x + 1} } } \cdot \dfrac{1}{\sqrt{\dfrac{x - 1}{x + 1} } } =[/tex]

[tex]= \dfrac{\dfrac{(x - 1)'(x + 1) - (x + 1)'(x - 1)}{(x + 1)^{2}} }{\dfrac{2(x - 1)}{(x + 1)} } = \dfrac{(x + 1)(x + 1 - (x - 1))}{2(x - 1)(x + 1)^{2}} = \dfrac{x+1-x +1}{2(x - 1)(x + 1)} =[/tex]

[tex]= \dfrac{2}{2(x - 1)(x + 1)} = \dfrac{1}{(x - 1)(x + 1)}=\dfrac{1}{x^{2} -1}[/tex]

2) [tex]f(x) = \sin^{3}7x[/tex]

[tex]f'(x) = (\sin^{3}7x)' = (\sin 7x)'\cdot 3\sin^{2} 7x = (7x)'\cdot 3 \cos 7x \sin^{2} 7x=[/tex]

[tex]= 7 \cdot 3 \cos 7x \sin^{2} 7x=21 \cos 7x \sin^{2} 7x[/tex]

3) [tex]f(x) = \ln \cos^{2} 8x[/tex]

[tex]f'(x) = (\ln \cos^{2} 8x)' = \dfrac{(\cos^{2} 8x)'}{\cos^{2} 8x} =\dfrac{(\cos 8x)' \cdot 2 \cos 8x}{\cos^{2} 8x} = -\dfrac{(8x)' \cdot 2 \sin 8x }{\cos 8x} =[/tex][tex]= -8 \cdot 2 \ \text{tg} \ 8x = -16\ \text{tg} \ 8x[/tex]