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Giải Toán 11 Bài 3: Đạo hàm của hàm số lượng giác

Bài tập 4 trang 169 SGK Toán lớp 11 Đại số:

a) y = (9 – 2x)(2x³ – 9x² + 1);

b) $y=\left(6\sqrt{x}-\frac{1}{x^2}\right)\left(7x-3\right)$;

c) $y=\left(x-2\right)\sqrt{x^2+1}$;

d) y = tan²x – cotx²;

e) $y=\cos \frac{x}{1+x}$.

Lời giải:

a) y = (9 – 2x)(2x³ – 9x² + 1)

$\Rightarrow$ y′ = (9 – 2x)′(2x³ – 9x² + 1) + (9 – 2x)(2x³ – 9x² + 1)′

= –2(2x³ – 9x² + 1) + (9 – 2x)(6x² – 18x)

= –4x³ +18x² – 2 + 54x² – 162x – 12x³ + 36x²

= –16x³ + 108x² – 162x – 2

Vậy y′ = –16x³ + 108x² – 162x – 2.

b) $y=\left(6\sqrt{x}-\frac{1}{x^2}\right)\left(7x-3\right)$

$\Rightarrow y^{\text{'}}={\left(6\sqrt{x}-\frac{1}{x^2}\right)}^{\text{'}}(7x-3)+\left(6\sqrt{x}-\frac{1}{x^2}\right)(7x-3{)}^{\text{'}}=\left(6\cdot \frac{1}{2\sqrt{x}}-\frac{-{\left(x^2\right)}^{\text{'}}}{{\left(x^2\right)}^2}\right)(7x-3)+\left(6\sqrt{x}-\frac{1}{x^2}\right).7=\left(\frac{3}{\sqrt{x}}+\frac{2x}{x^4}\right)(7x-3)+7\left(6\sqrt{x}-\frac{1}{x^2}\right)=\left(\frac{3}{\sqrt{x}}+\frac{2}{x^3}\right)(7x-3)+7\left(6\sqrt{x}-\frac{1}{x^2}\right)=21\sqrt{x}-\frac{9}{\sqrt{x}}+\frac{14}{x^2}-\frac{6}{x^3}+42\sqrt{x}-\frac{7}{x^2}=\frac{-6}{x^3}+\frac{7}{x^2}+63\sqrt{x}-\frac{9}{\sqrt{x}}$

Vậy $y^{\text{'}}=\frac{-6}{x^3}+\frac{7}{x^2}+63\sqrt{x}-\frac{9}{\sqrt{x}}$.

c) $y=\left(x-2\right)\sqrt{x^2+1}$

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

Vậy $y^{\text{'}}=\frac{2x^2-2x+1}{\sqrt{x^2+1}}$

d) y = tan²x – cotx²

$\Rightarrow$ y′ = (tan²x)′ – (cotx²)′

$=2tanx.{\left(tanx\right)}^{\text{'}}–{\left(x^2\right)}^{\text{'}}.\frac{-1}{{\sin }^2{x}^2}=2tanx.\frac{1}{{\cos }^{2}x}+\frac{2x}{{\sin }^2{x}^2}=\frac{2\sin x}{{\cos }^{3}x}+\frac{2x}{{\sin }^2{x}^2}$

Vậy $y^{\text{'}}=\frac{2\sin x}{{\cos }^{3}x}+\frac{2x}{{\sin }^2{x}^2}$.

e) $y=\cos \frac{x}{1+x}$

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

Vậy $y^{\text{'}}=-\frac{1}{{(1+x)}^2}.\sin \frac{x}{1+x}$.