xét hàm số y=\(\sqrt{x+\sqrt{x}}+\sqrt{x}\) . ta có

y'=\(\frac{\left(x+\sqrt{x}\right)}{2\sqrt{x+\sqrt{x}}}+\frac{1}{2\sqrt{x}}=\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}+\frac{1}{2\sqrt{x}}\)

=\(\frac{1+2\sqrt{x}}{4\sqrt{x}\sqrt{x+\sqrt{x}}}+\frac{1}{2\sqrt{x}}=\frac{1+2\sqrt{x}+2\sqrt{x+\sqrt{x}}}{4\sqrt{x}\sqrt{x+\sqrt{x}}}\)

a) y' = 2x - = 2x - .

b) y' = = .

c) y' = = = = .

d) y' = = = = .

Lời giải:

a) y' = = , y" = = = .

b) y' = = ;

y" = = = .

c) y' = ; y" = = = .

d) y' = 2cosx.(cosx)' = 2cosx.(-sinx) = - 2sinx.cosx = -sin2x,

y" = -(2x)'.cos2x = -2cos2x.

a) y' = 5cosx -3(-sinx) = 5cosx + 3sinx;

b) = = .

c) y' = cotx +x. = cotx -.

d) + = = (x. cosx -sinx).

e) = = .

f) y' = (√(1+x²))' cos√(1+x²) = cos√(1+x²) = cos√(1+x²).

Áp dụng công thức \(\left(\sqrt[n]{u}\right)'=\frac{u'}{n\sqrt[n]{u^{n-1}}}\) :

\(y'=\frac{1+\frac{1}{2\sqrt{x}}}{3\sqrt[3]{\left(x+\sqrt{x}\right)^2}}=\frac{2\sqrt{x}+1}{6\sqrt{x}\sqrt[3]{\left(x+\sqrt{x}\right)^2}}\)

\(y=\sqrt[3]{x+\sqrt{x}}\)

chia đều cho hai bên

\(\Leftrightarrow y^3=x+x^{\dfrac{1}{2}}\)

đạo hàm cấp 1{hai vế}

\(3y^2=1+\dfrac{1}{2}x^{\left(1-\dfrac{1}{2}\right)}=1+\dfrac{1}{2x^2}=\dfrac{2x^2+1}{2x^2}\)

Thay y=\(\sqrt[3]{x+\sqrt{x}}\) vào VT

\(\Leftrightarrow y'=\dfrac{\left(2x^2+1\right)}{2x^2.3.\sqrt[3]{\left(x+\sqrt{x}\right)^2}}=\dfrac{\left(2x^2+1\right)}{6.x^2.\sqrt[3]{\left(x+\sqrt{x}\right)^2}}\\ \)

Liệu có sai; --> sai ở đâu

xét hàm số y=ln(\(x+\sqrt{1+x^2}\))

Ta có

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

a) Cách 1: y' = (9 -2x)'(2x³- 9x² +1) +(9 -2x)(2x³- 9x² +1)' = -2(2x³- 9x² +1) +(9 -2x)(6x² -18x) = -16x³ +108x² -162x -2.

Cách 2: y = -4x⁴ +36x³ -81x² -2x +9, do đó

y' = -16x³ +108x² -162x -2.

b) y' = .(7x -3) +(7x -3)'= (7x -3) +7.

c) y' = (x -2)'√(x² +1) + (x -2)(√x² +1)' = √(x² +1) + (x -2) = √(x² +1) + (x -2) = √(x² +1) + = .

d) y' = 2tanx.(tanx)' - (x²)' = .

e) y' = sin = sin.