помогите решить производные [tex]1)y(x)= \frac{sinx}{cosx} \\ 2)y(x)=sin( \frac{5 \pi }{3} x) \\ 3)y(x)= \sqrt{x} - \frac{3}{x}+ \frac{9}{x^2} \\ 4)y(x)= \frac{x^2+4}{x^2-4} \\ 5)f(x)= \frac{ \sqrt{x} }{ \sqrt{x} +1} [/tex]. Created by natsysan1. algebra-ru

tex]...

Artem112

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$y= \cfrac{\sin x}{\cos x} 
\\\
y'= \cfrac{(\sin x)'\cdot\cos x-\sin x\cdot(\cos x)'}{\cos^2x} =
\\\
= \cfrac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^2x} =
 \cfrac{\cos^2 x+\sin ^2x}{\cos^2x} = \frac{1}{\cos^2x}$

$y=\sin( \frac{ 5\pi }{3} x)
\\\
y'=\cos( \frac{ 5\pi }{3} x)\cdot( \frac{ 5\pi }{3} x)'= \frac{ 5\pi }{3}\cos( \frac{ 5\pi }{3} x)$

$y= \sqrt{x} - \frac{3}{x}+ \frac{9}{x^2}=x^ \frac{1}{2} - 3x^{-1}+9x^{-2}
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y'= \frac{1}{2}x^ {\frac{1}{2}-1}-(- 3x^{-1-1})+9\cdot(-2x^{-2-1})= 
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=\frac{1}{2}x^ {-\frac{1}{2}}+ 3x^{-2}-18x^{-3}= \frac{1}{2 \sqrt{x} } + \frac{3}{x^2} - \frac{18}{x^3}$

$y= \cfrac{x^2+4}{x^2-4}
\\\
y= \cfrac{(x^2+4)'(x^2-4)-(x^2+4)(x^2-4)'}{(x^2-4)^2}=
\cfrac{2x(x^2-4)-2x(x^2+4)}{(x^2-4)^2}=
\\\
=\cfrac{2x(x^2-4-x^2-4)}{(x^2-4)^2}=\cfrac{2x\cdot(-8)}{(x^2-4)^2}=- \cfrac{16x}{(x^2-4)^2}$

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

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