Chương 5: ĐẠO HÀM

ta có :

y^' = -3x² -4x +1 → y^" = -6x -4 = 0

→ x = \(\dfrac{-2}{3}\)

Đáp án: C

c

C.

f^'(x) = 3x² - 6x

-> f^'(3) = 3.3³ - 6.3 = 9 

đáp án D

\( f'(x)=(x^3-3x^2+1)'=(x^3)'-(3x^2)'+(1)'=3x^2-6x\)

Đạo hàm của f(x) tại x=3 là

\(f'(3)=3.3^2-6.3=9\\\to D\)

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

→ y^'(1) = -7

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

\(x_0=2\Rightarrow y_0=2\Rightarrow y'\left(2\right)=\dfrac{2-1}{\sqrt{2^2-2.2+4}}=\dfrac{1}{2}\)

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

x_o = 2 → y_o = 2 

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

phương trình tiếp tuyến có dạng: 

y = \(\dfrac{1}{2}\)x +1

\(y^`=5\cdot sin^4\left(\sqrt{x^2+3x}\right)\cdot\left[sin\left(\sqrt{x^2+3x}\right)\right]^`\)

\(=5\cdot sin^4\left(\sqrt{x^2+3x}\right)\cdot cos\sqrt{x^2+3x}\)

\(y'=5\sin^4\left(\sqrt{x^2+3x}\right).\left[\sin\left(\sqrt{x^2+3x}\right)\right]'\)

\(=5\sin^4\left(\sqrt{x^2+3x}\right).\cos\left(\sqrt{x^2+3x}\right).\left(\sqrt{x^2+3x}\right)'\)

\(=5\sin^4\left(\sqrt{x^2+3x}\right)\cos\left(\sqrt{x^2+3x}\right).\dfrac{1}{2\sqrt{x^2+3x}}\left(x^2+3x\right)'\)

\(=5\sin^4\left(\sqrt{x^2+3x}\right)\cos\left(\sqrt{x^2+3x}\right).\dfrac{2x+3}{2\sqrt{x^2+3x}}\)

y^' = \(\dfrac{cos3x+3sin3x.x}{cos^23x}\) 

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