Question

Tính các đạo hàm của hàm số sau:
a) \(y=\sqrt{x}\left(x+3\right)\)

b) \(y=\sqrt{2x^2-6x-9}\)

c) \(y=\left(\sqrt{x^2+1}+x\right)^{10}\)

 

Answer 1

a: \(y=\sqrt{x}\left(x+3\right)\)

=>\(y^{\prime}=\sqrt{x}\cdot\left(x+3\right)^{\prime}+\left(\sqrt{x}\right)^{\prime}\cdot\left(x+3\right)=\sqrt{x}\cdot1+\frac{1}{2\sqrt{x}}\cdot x+3=\frac{x+3}{2\sqrt{x}}+\sqrt{x}=\frac{x+3+2x}{2\sqrt{x}}=\frac{3x+3}{2\sqrt{x}}\)

b: \(y=\sqrt{2x^2-6x-9}\)

=>\(y^{\prime}=\frac{\left(2x^2-6x-9\right)^{\prime}}{2\cdot\sqrt{2x^2-6x-9}}=\frac{4x-6}{2\cdot\sqrt{2x^2-6x-9}}=\frac{2x-3}{\sqrt{2x^2-6x-9}}\)