ta có :
y' = -3x2 -4x +1 → y" = -6x -4 = 0
→ x = \(\dfrac{-2}{3}\)
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
cho ham so y=\(\sqrt{x^2-2x+4}\) co đồ thị (C) . viết PTTT của (C) tại M có hoành độ \(x_o\)=2
\(y'=\dfrac{1}{2\sqrt{x^2-2x+4}}\left(x^2-2x+4\right)=\dfrac{1}{2\sqrt{x^2-2x+4}}.\left(2x-2\right)=\dfrac{x-1}{\sqrt{x^2-2x+4}}\)
\(x_0=2\Rightarrow y_0=2\Rightarrow y'\left(2\right)=\dfrac{2-1}{\sqrt{2^2-2.2+4}}=\dfrac{1}{2}\)
\(y=\dfrac{1}{2}\left(x-2\right)+2\)
xo = 2 → yo = 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
\(y^`=5\cdot sin^4\left(\sqrt{x^2+3x}\right)\cdot\left[sin\left(\sqrt{x^2+3x}\right)\right]^`\)
\(=5\cdot sin^4\left(\sqrt{x^2+3x}\right)\cdot cos\sqrt{x^2+3x}\)
\(y'=5\sin^4\left(\sqrt{x^2+3x}\right).\left[\sin\left(\sqrt{x^2+3x}\right)\right]'\)
\(=5\sin^4\left(\sqrt{x^2+3x}\right).\cos\left(\sqrt{x^2+3x}\right).\left(\sqrt{x^2+3x}\right)'\)
\(=5\sin^4\left(\sqrt{x^2+3x}\right)\cos\left(\sqrt{x^2+3x}\right).\dfrac{1}{2\sqrt{x^2+3x}}\left(x^2+3x\right)'\)
\(=5\sin^4\left(\sqrt{x^2+3x}\right)\cos\left(\sqrt{x^2+3x}\right).\dfrac{2x+3}{2\sqrt{x^2+3x}}\)
s ra đc cos2x vậy ??