Question

Answer 1

Câu 1:

Ta có: \(y'=2x^3-\dfrac{1}{x^2}+\dfrac{1}{\sqrt{2x}}\)

\(\Rightarrow y'\left(2\right)=\dfrac{65}{4}\)

Câu 2:

Ta có: \(y'=2sinx.cosx-2sin2x=sin2x-2sin2x=-sin2x\)

\(\Rightarrow y'\left(\dfrac{\pi}{6}\right)=\dfrac{-\sqrt{3}}{2}\)

Câu 3:

Ta có: x₀ = 2 ⇒ y₀ = f(2) = -5

\(y'=3x^2-6x+2\Rightarrow y'\left(2\right)=2\)

→ PTTT cần tìm là: y = 2(x - 2) - 5 = 2x - 9

Answer 2

1.

\(y'=\left(\dfrac{x^4}{2}+\dfrac{1}{x}+\sqrt{2x}\right)'\)

\(=\left(\dfrac{x^4}{2}\right)'+\left(\dfrac{1}{x}\right)'+\left(\sqrt{2x}\right)'\)

\(=2x^3-\dfrac{1}{x^2}+\dfrac{\sqrt{2}}{2\sqrt{x}}\)

\(\Rightarrow y'\left(2\right)=16-\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{65}{4}\)