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Giải Toán 11 Bài 5: Đạo hàm cấp 2

Bài tập 2 trang 174 SGK Toán lớp 11 Đại số:

a) $y=\frac{1}{1-x}$;

b) $y=\frac{1}{\sqrt{1-x}}$;

c) y = tanx ;

d) y = cos²x.

Lời giải:

a) $y=\frac{1}{1-x}$

$\begin{array}{l}\Rightarrow y^{\text{'}}=\frac{-(1-x{)}^{\text{'}}}{{(1-x)}^2}=\frac{-(-1)}{{(1-x)}^2}=\frac{1}{{(1-x)}^2}\\ \Rightarrow y^{\text{'}\text{'}}=-\frac{{\left.{(1-x)}^2\right]}^{\text{'}}}{{(1-x)}^4}=-\frac{2(1-x)(-1)}{{(1-x)}^4}=\frac{2}{{(1-x)}^3}.\end{array}$

b) $y=\frac{1}{\sqrt{1-x}}$

$\begin{array}{l}\Rightarrow y^{\text{'}}=-\frac{{\left(\sqrt{1-x}\right)}^{\text{'}}}{{\left(\sqrt{1-x}\right)}^2}=-\frac{\frac{(1-x{)}^{\text{'}}}{2\sqrt{1-x}}}{1-x}=-\frac{-1}{2\sqrt{1-x}}=\frac{1}{2{\left(\sqrt{1-x}\right)}^3}\\ \Rightarrow y^{\text{'}\text{'}}=\frac{1}{2}\cdot \frac{-{\left.{\left(\sqrt{1-x}\right)}^3\right]}^{\text{'}}}{{\left(\sqrt{1-x}\right)}^6}=-\frac{1}{2}\cdot \frac{3{\left(\sqrt{1-x}\right)}^2.{\left(\sqrt{1-x}\right)}^{\text{'}}}{{\left(\sqrt{1-x}\right)}^6}\\ =-\frac{3(1-x)\cdot \frac{-1}{2\sqrt{1-x}}}{2{\left(\sqrt{1-x}\right)}^6}=\frac{3}{4{\left(\sqrt{1-x}\right)}^5}\end{array}$

c) y = tanx

$\begin{array}{l}\Rightarrow y^{\text{'}}=\frac{1}{{\cos }^{2}x}\\ \Rightarrow y^{\text{'}\text{'}}=-\frac{{\left({\cos }^{2}x\right)}^{\text{'}}}{{\cos }^{4}x}=-\frac{2\cos x(\cos x{)}^{\text{'}}}{{\cos }^{4}x}=\frac{2\cos x\sin x}{{\cos }^{4}x}=\frac{2\sin x}{{\cos }^{3}x}\end{array}$

Cách khác:

$y^{\text{'}}=\frac{1}{{\cos }^{2}x}=1+{\tan }^{2}x$

y’’ = (1+ tan² x)’ = 2tan x(tan x)’ $=2\tan x\cdot \frac{1}{{\cos }^{2}x}=\frac{2\tan x}{{\cos }^{2}x}=\frac{2\sin x}{{\cos }^{3}x}$

d) y = cos²x

$\Rightarrow$ y′ = 2cosx(cosx)′ = –2cosxsinx = – sin2x

$\Rightarrow$ y′′ = –(2x)′cos2x = –2cos2x.