1.
[tex]y=-3tgx+x^3+x[/tex]
[tex]y'=(-3tgx+x^3+x)'=-\frac{3}{cos^2x} +3x^{2} +1[/tex]
[tex]y'=-\frac{3}{cos^2x} +3x^{2} +1[/tex]
[tex]y'(0)=-\frac{3}{cos^20} +3*0^{2} +1=-\frac{3}{1} +0+1=-3+1=-2[/tex]
[tex]y'(0)=-2[/tex]
1 => Д
2.
[tex]y=\frac{x}{sinx}[/tex]
[tex]y'=(\frac{x}{sinx} )'=\frac{x'sinx-x(sinx)'}{sin^2x} =\frac{sinx-xcosx}{sin^2x}[/tex]
[tex]y'=\frac{sinx-xcosx}{sin^2x}[/tex]
[tex]y'(\frac{\pi }{2} )=\frac{sin\frac{\pi }{2} -\frac{\pi }{2} *cos\frac{\pi }{2} }{sin^2\frac{\pi }{2} } =\frac{1-\frac{\pi }{2} *0}{1^2}=\frac{1}{1}=1[/tex]
[tex]y'(\frac{\pi }{2} )=1[/tex]
2 => А
3.
[tex]y=4\sqrt{x} -x^3+1[/tex]
[tex]y'=(4\sqrt{x} -x^3+1)'=\frac{4}{2\sqrt{x} } -3x^{2} +0=\frac{2}{\sqrt{x} } -3x^{2}[/tex]
[tex]y'=\frac{2}{\sqrt{x} } -3x^{2}[/tex]
[tex]y'(1)=\frac{2}{\sqrt{1} } -3*1^{2} =2-3=-1[/tex]
[tex]y'(1)=-1[/tex]
3 => В
Ответ: 1 => Д
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Verified answer
1.
[tex]y=-3tgx+x^3+x[/tex]
[tex]y'=(-3tgx+x^3+x)'=-\frac{3}{cos^2x} +3x^{2} +1[/tex]
[tex]y'=-\frac{3}{cos^2x} +3x^{2} +1[/tex]
[tex]y'(0)=-\frac{3}{cos^20} +3*0^{2} +1=-\frac{3}{1} +0+1=-3+1=-2[/tex]
[tex]y'(0)=-2[/tex]
1 => Д
2.
[tex]y=\frac{x}{sinx}[/tex]
[tex]y'=(\frac{x}{sinx} )'=\frac{x'sinx-x(sinx)'}{sin^2x} =\frac{sinx-xcosx}{sin^2x}[/tex]
[tex]y'=\frac{sinx-xcosx}{sin^2x}[/tex]
[tex]y'(\frac{\pi }{2} )=\frac{sin\frac{\pi }{2} -\frac{\pi }{2} *cos\frac{\pi }{2} }{sin^2\frac{\pi }{2} } =\frac{1-\frac{\pi }{2} *0}{1^2}=\frac{1}{1}=1[/tex]
[tex]y'(\frac{\pi }{2} )=1[/tex]
2 => А
3.
[tex]y=4\sqrt{x} -x^3+1[/tex]
[tex]y'=(4\sqrt{x} -x^3+1)'=\frac{4}{2\sqrt{x} } -3x^{2} +0=\frac{2}{\sqrt{x} } -3x^{2}[/tex]
[tex]y'=\frac{2}{\sqrt{x} } -3x^{2}[/tex]
[tex]y'(1)=\frac{2}{\sqrt{1} } -3*1^{2} =2-3=-1[/tex]
[tex]y'(1)=-1[/tex]
3 => В
Ответ: 1 => Д
2 => А
3 => В