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Giải Toán 11 Bài 2: Quy tắc tính đạo hàm

Bài tập 2 trang 163 SGK Toán lớp 11 Đại số:

a) y = x⁵ – 4x³ + 2x – 3;

b) $y=\frac{1}{4}-\frac{1}{3}x+x^2-0,5x^4$;

c) $y=\frac{x^4}{2}-\frac{2x^3}{3}+\frac{4x^2}{5}-1$;

d) y = 3x⁵(8 – 3x²).

Lời giải:

a) y′ = (x⁵ − 4x³ + 2x − 3)′

= (x⁵)′ − (4x³)′ + (2x)′ − (3)′

= (x⁵)′ − 4.(x³)′ + 2.(x)′ − 0

= 5x⁴ − 4.3x² + 2

= 5x⁴ −12x² + 2b)

$y\text{'}={\left(\frac{1}{4}-\frac{1}{3}x+x^2-0,5x^4\right)}^{\text{'}}$

$={\left(\frac{1}{4}\right)}^{\text{'}}-{\left(\frac{1}{3}x\right)}^{\text{'}}+{\left(x^2\right)}^{\text{'}}-{\left(0,5x^4\right)}^{\text{'}}$

$=0-\frac{1}{3}\left(x\right)\text{'}+\left(x^2\right)\text{'}-0,5\left(x^4\right)\text{'}$

$=-\frac{1}{3}+2x-0,5.4x^3$

$=-\frac{1}{3}+2x-2x^3$

c)

$y^{\text{'}}={\left(\frac{x^4}{2}-\frac{2x^3}{3}+\frac{4x^2}{5}-1\right)}^{\text{'}}$

$={\left(\frac{x^4}{2}\right)}^{\text{'}}-{\left(\frac{2x^3}{3}\right)}^{\text{'}}+{\left(\frac{4x^2}{5}\right)}^{\text{'}}-1^{\text{'}}$

$=\frac{1}{2}{\left(x^4\right)}^{\text{'}}-\frac{2}{3}{\left(x^3\right)}^{\text{'}}+\frac{4}{5}{\left(x^2\right)}^{\text{'}}-0$

$=\frac{1}{2}.4x^3-\frac{2}{3}.3x^2+\frac{4}{5}.2x$

$=2x^3-2x^2+\frac{8}{5}x$

d) y = 3x⁵(8 – 3x²) = 24x⁵ – 9x⁷

y′ = (24x⁵ – 9x⁷)′

= 24(x⁵)′ – 9(x⁷)′

= 24.5x⁴ – 9.7x⁶

=120x⁴ – 63x⁶

Cách khác:

y′ = [3x⁵(8 – 3x²)]′

= (3x⁵)′(8 – 3x²) + 3x⁵(8 – 3x²)′

= 3(x⁵)′(8 – 3x²) + 3x⁵ [8′ – (3x²)′]

= 3.5x⁴ (8 – 3x²) + 3x⁵ (0 – 3.2x)

= 120x⁴ – 45x⁶ – 18x⁶

= 120x⁴ – 63x⁶