1 ) y = 1 2 ( 3 x 2 + 1 ) 2 y ′ = 1 2 ( − 2 ) 1 ( 3 x 2 + 1 ) 3 ⋅ ( 6 x ) = = − 6 x ( 3 x 2 + 1 ) 2 2 ) y = 3 + 4 x − x 2 y ′ = 0 + 4 − 2 x = 4 − 2 x 3 ) y = x 2 − 2 2 x 3 + 1 y ′ = 1 2 x 2 − 2 ⋅ 2 x ⋅ ( 2 x 3 + 1 ) − x 2 − 2 ⋅ 6 x 2 ( 2 x 3 + 1 ) 2 = = x ⋅ ( 2 x 3 + 1 ) − ( x 2 − 2 ) ⋅ 6 x 2 x 2 − 2 ( 2 x 3 + 1 ) 2 = = − 4 x 4 + 12 x 2 + x x 2 − 2 ( 2 x 3 + 1 ) 2 4 ) y = x 2 6 x 2 3 x y = x 6 x 2 3 = 6 3 x 2 y ′ = 6 3 2 x = 2 6 x 3 5 ) y = 2 x 2 x = 2 x 5 2 y ′ = 2 ⋅ 5 2 ⋅ x 3 2 = 5 x 3 2 1) y=\frac{1}{2(3x^2+1)^2}\\
y'=\frac{1}{2}(-2)\frac{1}{(3x^2+1)^3}\cdot(6x)=\\
=\frac{-6x}{(3x^2+1)^2}\\
2) y=3+4x-x^2\\
y'=0+4-2x=4-2x\\
3) y=\frac{\sqrt{x^2-2}}{2x^3+1}\\
y'=\frac{\frac{1}{2\sqrt{x^2-2}}\cdot 2x\cdot (2x^3+1)- \sqrt{x^2-2}\cdot 6x^2}{(2x^3+1)^2}=\\
=\frac{x\cdot (2x^3+1)- (x^2-2)\cdot 6x^2}{\sqrt{x^2-2}(2x^3+1)^2}=\\
=\frac{-4x^4+12x^2+x}{\sqrt{x^2-2}(2x^3+1)^2}\\
4) y=\frac{x^2\sqrt{6x^2}}{3x}\\
y=\frac{x\sqrt{6x^2}}{3}=\frac{\sqrt{6}}{3}x^2\\
y'=\frac{\sqrt{6}}{3}2x=\frac{2\sqrt{6}x}{3}\\
5) y=2x^2\sqrt{x}=2x^{\frac{5}{2}}\\
y'=2\cdot\frac{5}{2}\cdot x^{\frac{3}{2}}=5x^{\frac{3}{2}} 1 ) y = 2 ( 3 x 2 + 1 ) 2 1 y ′ = 2 1 ( − 2 ) ( 3 x 2 + 1 ) 3 1 ⋅ ( 6 x ) = = ( 3 x 2 + 1 ) 2 − 6 x 2 ) y = 3 + 4 x − x 2 y ′ = 0 + 4 − 2 x = 4 − 2 x 3 ) y = 2 x 3 + 1 x 2 − 2 y ′ = ( 2 x 3 + 1 ) 2 2 x 2 − 2 1 ⋅ 2 x ⋅ ( 2 x 3 + 1 ) − x 2 − 2 ⋅ 6 x 2 = = x 2 − 2 ( 2 x 3 + 1 ) 2 x ⋅ ( 2 x 3 + 1 ) − ( x 2 − 2 ) ⋅ 6 x 2 = = x 2 − 2 ( 2 x 3 + 1 ) 2 − 4 x 4 + 12 x 2 + x 4 ) y = 3 x x 2 6 x 2 y = 3 x 6 x 2 = 3 6 x 2 y ′ = 3 6 2 x = 3 2 6 x 5 ) y = 2 x 2 x = 2 x 2 5 y ′ = 2 ⋅ 2 5 ⋅ x 2 3 = 5 x 2 3
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