3a-2a+4+a-1=2a+3.Ответ: 2a+3.
Где зачеркнуто там сокращается
1.
а) f(x) =(1/2)*x⁴ -x⁵ +5 ;
f '(x) = ((1/2)*x⁴ -x⁵ +5)= (1/2)*4*x³ -5x⁴ +0 =2x³ -5x⁴.
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б)<span> f(x)=4x -1/x³.
</span> f'(x)=(4x -1/x³) ' =<span>(4x - x^(-3)</span><span>) '= </span>4 + 3/x⁴<span>.
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</span><span>2.
</span>а) f(x)=x*cosx
f'(x)= (x*cosx)' =(x)'*cosx+x*(cosx)' =<span>cosx-x*sinx.
</span>f'(-π/2)= cos(-π/2)- (-π/2)*sin(-π/2) =cosπ/2 +π/2*(-sinπ/2)= - π/2.
б) f(x) =(3x+4)⁵ ;
f'(x) = ((3x+4)⁵)' =5(3x+4)⁴*(3x+4)'=5(3x+4)⁴*3=15(3x+4)⁴<span>.
</span>f'(-1) =15(3*(-1)+4)⁴ =15<span>.
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</span><span>3.
</span>f(x) =cos2x +√3*x ; f '(x)=0 . ||ошибочно f <span>'(-1)=0 ???? ||</span>
f '(x) =(cos2x +√3*x)' = -2sin2x +√3 .
f '(x) =0 ⇔ -2sin2x +√3 =0⇒sin2x =(<span>√3)/2 ;
</span>2x =(-1)^n*π/3 +πn , n∈Z.
x =(-1)^n*π<span>/6 +π*n/2 , n∈Z.
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</span><span>4.
</span>f(x) =6x -x³ ; f '(x) ≤0.
f '(x) =(6x -x³)' =6 -3x² =3(2 -x²).
f'(x) ≤ 0 ⇒ 3(2 -x²) ≤ 0 ⇔ x² -2 ≥0 ⇒x∈(-∞; -√2] ∪[√2;∞).
При а=0
то х будет равнятся = -1