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' = = = = .

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 sau       \(\frac{x+\sqrt{x^2+1}}{\sqrt{1+x^2-x}}+\frac{\sqrt{1+x^2-x}}{x+\sqrt{x^2+1}}\)

=\(\frac{\left(x+\sqrt{x^2+1}\right)\left(\sqrt{1+x^2}+x\right)}{\left(1+x^2\right)-x^2}+\frac{\left(\sqrt{1+x^2-x}\right)\left(\sqrt{x^2+1}-x\right)}{x^2+1-x^2}=\left(x+\sqrt{x^2+1}\right)^2+\left(\sqrt{x^2+1-x}\right)^2=4x^2+2\)