Đáp án:

Câu `1:`

`1) y = log_2 (3x^2-6x)`

`y' = ((3x^2-6x)')/((3x^2-6x).ln2)`

`= (6x-6)/((3x^2-6x).ln2)`

`2) y = 1/(4x^3-1)`

`y' = (1'.(4x^3-1)-(4x^3-1)'.1)/((4x^3-1)^2)`

`= (-12x^2)/((4x^3-1)^2)`

`3) y = 10^(3x^4+6x)`

`y' = (3x^4+6x)' .10^(3x^4+6x) .ln10`

`= (12x^3+6).ln10 .10^(3x^4+6x)`

`4) y = e^(log_6 3x^2)`

`y' = (log_6 3x^2)' .e^(log_6 3x^2)`

`= ((3x^2)')/(3x^2 .ln6) .e^(log_6 3x^2)`

`= 2/(x.ln6) .e^(log_6 3x^2)`

Câu `2:`

`1) y = 11/(x^4) +6x^5 -7x^3`

`y' = (-(x^4)'.11)/(x^8) +30x^4 -21x^2`

`= (-44)/(x^5) +30x^4 -21x^2`

`y'' = (-(x^5)'.(-44))/(x^10) +120x^3 -42x`

`= (220)/(x^6) +120x^3 -42x`

`y''(1) = 220/1 +120.1-42.1 = 298`

`2) y = \sqrt{x+1}`

`y' = ((x+1)')/(2\sqrt{x+1}) = 1/(2\sqrt{x+1)}`

`y'' = (-(2\sqrt{x+1})')/(4.(x+1))`

`= (-2. 1/(2\sqrt{x+1}))/(4x+4)`

`= (-\sqrt{x+1})/(4x+4)`

`y''(1) = (-\sqrt{1+1})/(4.1+4) = (-\sqrt{2})/8`

`3) y = 2^(x^4-5x)`

`y' = (x^4-5x)'.2^(x^4-5x) .ln2`

`= (4x^3-5) .2^(x^4-5x) .ln2`

`y'' = ln2. [(4x^3-5)' .2^(x^4-5x) + [2^(x^4-5x)]' .(4x^3-5)]`

`= ln2 .(12x^2-5) .2^(x^4-5x) + ln2 .(4x^3-5) .2^(x^4-5x) .ln2 .(4x^3-5)`

`y''(1) = ln2. (12.1-5) .2^(1-5) +ln2 .(4.1-5) .2^(1-5) .ln2 .(4.1-5) ~~ 0,33`