1/7x+3/14x=14Ответь те плииз. =3
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<span>Найти производные следующих функций
1) y = 3 – 2x + (2/3)x</span>⁴; 2) y = (x−1)/(x+1); 3) y = x³·ctgx; 4) y = cosx/(1+sinx); 5) у = (x²-1) ·(x²-4) ·(x²<span>+9); 6) y = 2sinx – 3tgx;
Решение:
</span>1) y' = (3 – 2x + (2/3)x⁴)' = (3)' – (2x)' + ((2/3)x⁴)' = 0 - 2 + (2/3)·4x⁴⁻¹=
= - 2 + 8x³/3
2)
3) y' = (x³·ctgx)' =(x³)'·ctgx + x³·(ctgx)' = 3x²·ctgx - x³/sin²x
4)
5) у' = (x²-1)' ·(x²-4) ·(x²+9)+ (x²-1) ·(x²-4)' ·(x²+9)+ (x²-1)' ·(x²-4) ·(x²+9)'=
= 2x·(x²-4) ·(x²+9) + (x²-1) ·2x·(x²+9) + (x²-1)·(x²-4) ·2x =
= 2x·((x²-4) ·(x²+9) + (x²-1)·(x²+9) + (x²-1)·(x²-4))
6) y' = (2sinx – 3tgx)' = (2sinx)' – (3tgx)' = 2cosx - 3/cos²x
1) y = 3 – 2x + (2/3)x</span>⁴; 2) y = (x−1)/(x+1); 3) y = x³·ctgx; 4) y = cosx/(1+sinx); 5) у = (x²-1) ·(x²-4) ·(x²<span>+9); 6) y = 2sinx – 3tgx;
Решение:
</span>1) y' = (3 – 2x + (2/3)x⁴)' = (3)' – (2x)' + ((2/3)x⁴)' = 0 - 2 + (2/3)·4x⁴⁻¹=
= - 2 + 8x³/3
2)
3) y' = (x³·ctgx)' =(x³)'·ctgx + x³·(ctgx)' = 3x²·ctgx - x³/sin²x
4)
5) у' = (x²-1)' ·(x²-4) ·(x²+9)+ (x²-1) ·(x²-4)' ·(x²+9)+ (x²-1)' ·(x²-4) ·(x²+9)'=
= 2x·(x²-4) ·(x²+9) + (x²-1) ·2x·(x²+9) + (x²-1)·(x²-4) ·2x =
= 2x·((x²-4) ·(x²+9) + (x²-1)·(x²+9) + (x²-1)·(x²-4))
6) y' = (2sinx – 3tgx)' = (2sinx)' – (3tgx)' = 2cosx - 3/cos²x