Tìm đạo hàm của các hàm số sau y = (x7 – 5x2)3

Giải Toán 11 Bài 2: Quy tắc tính đạo hàm

Bài tập 3 trang 163 SGK Toán lớp 11 Đại số: Tìm đạo hàm của các hàm số sau:

a) y = (x⁷ – 5x²)³ ;

b) y = (x² + 1)(5 – 3x²) ;

c)$y=\frac{2x}{x^2-1}$ ;

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

e) $y={\left(m+\frac{n}{x^2}\right)}^3$ (m, n là các hằng số).

Lời giải:

a) Áp dụng công thức đạo hàm hàm hợp y = u³, u = x⁷ − 5x²

y = (x⁷ − 5x²)³

$\Rightarrow$y′ = 3(x⁷ − 5x²)² (x⁷ − 5x²)′

= 3(x⁷ − 5x²)² [(x⁷)′ − (5x²)′]

= 3(x⁷ − 5x2)² . (7x⁶ − 5.2x)

= 3(x⁷ − 5x²)² . (7x⁶ − 10x)

b) y = (x² + 1)(5 − 3x²)

$\Rightarrow$ y = 5x² − 3x⁴ + 5 − 3x²= −3x⁴ + 2x² + 5

$\Rightarrow$ y′ = (−3x⁴)′ + (2x²)′ + (5)′= −3.4x³ + 2.2x + 0= −12x³ + 4x

Cách khác:

y′ = (x² + 1)′ (5 − 3x²) + (x² + 1)(5 − 3x²)′

= [(x²)′ + (1)′](5 − 3x²) + (x² + 1)[(5)′ − (3x²)′]

= (2x + 0)(5 − 3x²) + (x² + 1)(0 − 3.2x)

= 10x − 6x³ − 6x³ − 6x

= 4x − 12x³

c)$y=\frac{2x}{x^2-1}$

$y\text{'}=\frac{(2x{)}^{\text{'}}\left(x^2-1\right)-2x\cdot {\left(x^2-1\right)}^{\text{'}}}{{\left(x^2-1\right)}^2}$$=\frac{2\left(x^2-1\right)-2x.2x}{{\left(x^2-1\right)}^2}=\frac{2x^2-2-4x^2}{{\left(x^2-1\right)}^2}=\frac{-2x^2-2}{{\left(x^2-1\right)}^2}$

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

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

$=\frac{-5\left(x^2-x+1\right)-(3-5x)(2x-1)}{{\left(x^2-x+1\right)}^2}=\frac{-5x^2+5x-5+3-11x+10x^2}{{\left(x^2-x+1\right)}^2}=\frac{5x^2-6x-2}{{\left(x^2-x+1\right)}^2}$

e)$y={\left(m+\frac{n}{x^2}\right)}^3$

$y^{\text{'}}=3{\left(m+\frac{n}{x^2}\right)}^2{\left(m+\frac{n}{x^2}\right)}^{\text{'}}=3{\left(m+\frac{n}{x^2}\right)}^2\left.(m{)}^{\text{'}}+{\left(\frac{n}{x^2}\right)}^{\text{'}}\right]=3{\left(m+\frac{n}{x^2}\right)}^2\left.0+\frac{(n{)}^{\text{'}}.x^2-n.{\left(x^2\right)}^{\text{'}}}{x^4}\right]=3{\left(m+\frac{n}{x^2}\right)}^2.\frac{0x^2-n.2x}{x^4}=3{\left(m+\frac{n}{x^2}\right)}^2.\frac{-2n}{x^3}=-6n{\left(m+\frac{n}{x^2}\right)}^2.\frac{1}{x^3}$

Cách khác:

$y={\left(m+\frac{n}{x^2}\right)}^3\Rightarrow y^{\text{'}}=3{\left(m+\frac{n}{x^2}\right)}^2{\left(m+\frac{n}{x^2}\right)}^{\text{'}}=3{\left(m+\frac{n}{x^2}\right)}^2.{\left(m+n.x^{-2}\right)}^{\text{'}}=3{\left(m+\frac{n}{x^2}\right)}^2.n.(-2).x^{-3}=-6n{\left(m+\frac{n}{x^2}\right)}^2.\frac{1}{x^3}$.