_BT_QUY_TAC_TINH_DAO_HAM.docx_366a38_09_12_2025

Ta có: $f^{\prime }\left(x\right)=x^2-4\sqrt{2}x+8$.

Phương trình $f^{\prime }\left(x\right)=0\iff x^2-4\sqrt{2}x+8=0\iff x=2\sqrt{2}$.

Ta có: $y^{\prime }=9x^2+2x$.

Do đó, $y^{\prime }\le 0\iff y^{\prime }=9x^2+2x\le 0\iff -\frac{2}{9}\le x\le 0x\in \left[-\frac{2}{9};0\right]$.

Ta có: f^{\prime }\left(x\right)=-4x^3+12x{}^2-6x+2.

Suy ra f^{\prime }\left(-1\right)=-4{\left(-1\right)}^3+12\left(-1\right){}^2-6\left(-1\right)+2=24.

Ta có: $y^{\prime }=x^2-2\left(2m+1\right)x-m$.

Khi đó, $y\text{'}\ge 0$ với \forall x\in ℝ $\iff x^2-2\left(2m+1\right)x-m\ge 0$ với \forall x\in ℝ

$\iff {\Delta }^{\prime }=\left(2m+1\right){}^2+m\le 0\iff 4m{}^2+5m+1\le 0\iff -1\le m\le -\frac{1}{4}$.

Ta có: $y^{\prime }=-m{x}^2+2\left(m-1\right)x-m$.

Phương trình $y^{\prime }=0$ có hai nghiệm phân biệt

\iff -m{x}^2+2\left(m-1\right)x-m=0 có 2 nghiệm phân biệt

$\iff \{\begin{array}{l}m\ne 0\\ {\Delta }^{\prime }=\left(m-1\right){}^2-m{}^2>0\end{array}\iff \{\begin{array}{l}m\ne 0\\ m<\frac{1}{2}\end{array}$ .

Khi đó, gọi $x_1,x{}_2$ là hai nghiệm phân biệt của phương trình $\Rightarrow \{\begin{array}{l}{x}_1+x{}_2=\frac{2\left(m-1\right)}{m}\\ x_{1}x{}_2=1\end{array}$ .

Ta có: $x_1^2+x_2^2=6\iff {\left(x_1+x{}_2\right)}^2-2x{}_1{x}_2=6\iff \left(\frac{2\left(m-1\right)}{m}\right){}^2-2=6$

$\iff m^2+2m-1=0\iff m=-1\pm \sqrt{2}$.

So với điều kiện thì $m=-1\pm \sqrt{2}$ thỏa yêu cầu bài toán.

Ta có $f^{\prime }\left(x\right)=3a{x}^2+2bx+c$. Vì $a>0$ và $f^{\prime }\left(x\right)>0$ với $\forall x\in ℝ$ nên ${\Delta }^{\prime }<0$ tức là b^2-3ac<0.

Ta có $f^{\prime }\left(x\right)=3a{x}^2+2bx+c$. Vì a<0 và $f^{\prime }\left(x\right)<0$ với $\forall x\in ℝ$ nên ${\Delta }^{\prime }<0$ tức là b^2-3ac<0.

Ta có: $y^{\prime }=2{\left(x^3-2x^2\right)}^{\prime }\left(x^3-2x^2\right)=2\left(3x^2-4x\right)\left(x^3-2x^2\right)=6x^5-20x^4+16x^3$.

Ta có: $y^{\prime }=3{\left(2x^2+1\right)}^{\prime }{\left(2x^2+1\right)}^2=3.4x{\left(2x^2+1\right)}^2=12x{\left(2x^2+1\right)}^2$.

Do đó, $y^{\prime }\ge 0\iff 12x{\left(2x^2+1\right)}^2\iff x\ge 0$.

Ta có: y^{\prime }=4{\left(7x-5\right)}^{\prime }{\left(7x-5\right)}^3=28{\left(7x-5\right)}^3=-28{\left(5-7x\right)}^3.

Ta có: y^{\prime }=5{\left(1-x^3\right)}^{\prime }{\left(1-x^3\right)}^4=5\left(-3x^2\right){\left(1-x^3\right)}^4=-15x^2{\left(1-x^3\right)}^4.

Ta có: y^{\prime }=2016{\left(x^3-2x^2\right)}^{\prime }{\left(x^3-2x^2\right)}^{2015}=2016\left(3x^2-4x\right){\left(x^3-2x^2\right)}^{2015}.

Ta có:$y^{\prime }={\left(x^2-2\right)}^{\prime }\left(2x-1\right)+\left(x^2-2\right){\left(2x-1\right)}^{\prime }=2x\left(2x-1\right)+2\left(x^2-2\right)=6x^2-2x-4$

Xét hàm số $f\left(x\right)=f_0\left(x\right)f_1\left(x\right)f_2\left(x\right)...f_n\left(x\right)\left(n\ge 1;n\in \mathbb{Z}\right)$.

Bằng quy nạp, dễ dàng chứng minh được:

$f^{\prime }\left(x\right)=f_0{}^{\prime }\left(x\right)f_1\left(x\right)...f_n\left(x\right)+f_0\left(x\right){f^{\prime }}_1\left(x\right)...f_n\left(x\right)+...+f_0\left(x\right)f_1\left(x\right)...{f^{\prime }}_n\left(x\right)$

Áp dụng công thức trên cho hàm số $f\left(x\right)=x\left(x-1\right)\left(x-2\right)...\left(x-2018\right)$ và thay $x=0$ với chú ý $f_0\left(0\right)=0$ ta được:

$f^{\prime }\left(0\right)=\left(-1\right).\left(-2\right)...\left(-2018\right)+0.\left(-2\right)....\left(-2018\right)+0.\left(-1\right)...\left(-2017\right)=2018!$.

Xét hàm số $f\left(x\right)=f_0\left(x\right)f_1\left(x\right)f_2\left(x\right)...f_n\left(x\right)\left(n\ge 1;n\in \mathbb{Z}\right)$.

Bằng quy nạp, dễ dàng chứng minh được:

$f^{\prime }\left(x\right)=f_0{}^{\prime }\left(x\right)f_1\left(x\right)...f_n\left(x\right)+f_0\left(x\right){f^{\prime }}_1\left(x\right)...f_n\left(x\right)+...+f_0\left(x\right)f_1\left(x\right)...{f^{\prime }}_n\left(x\right)$.

Áp dụng công thức trên cho hàm số $f\left(x\right)=x\left(x+1\right)\left(x+2\right)...\left(x+2018\right)$ và thay $x=-1004$ với chú ý $f_{1004}\left(-1004\right)=0$ ta được

$\begin{array}{l}{f}^{\prime }\left(-1004\right)=\left[\left(-1004\right).\left(-1004+1\right)...\left(-1004+1003\right)\right].\left[\left(-1004+1005\right)...\left(-1004+2018\right)\right]\\ =\left(-1\right).1.\left(-2\right).2.....\left(-1004\right).1004={\left(1004!\right)}^2.\end{array}$

TXĐ: $\text{D}=ℝ\backslash \left\{1\right\}.$

Ta có $f^{\prime }\left(x\right)=\frac{-2}{{\left(x-1\right)}^2}\Rightarrow f^{\prime }\left(-1\right)=-\frac{1}{2}$

Ta có $y=x-\frac{3}{x+2}\Rightarrow y^{\prime }=1+\frac{3}{{\left(x+2\right)}^2}$.

Ta có: y=\frac{x\left(1-3x\right)}{x+1}=\frac{x-3x^2}{x+1}

\Rightarrow y^{\prime }=\frac{{\left(x-3x^2\right)}^{\prime }\left(x+1\right)-\left(x-3x^2\right){\left(x+1\right)}^{\prime }}{{\left(x+1\right)}^2}=\frac{\left(1-6x\right)\left(x+1\right)-\left(x-3x^2\right)}{{\left(x+1\right)}^2}=\frac{-3x^2-6x+1}{{\left(x+1\right)}^2}.

Ta có $f^{\prime }\left(x\right)=\frac{{\left(x^2+x\right)}^{\prime }\left(x-2\right)-\left(x^2+x\right){\left(x-2\right)}^{\prime }}{{\left(x-2\right)}^2}$

$=\frac{\left(2x+1\right)\left(x-2\right)-\left(x^2+x\right)}{{\left(x-2\right)}^2}=\frac{x^2-4x-2}{{\left(x-2\right)}^2}\Rightarrow f^{\prime }\left(1\right)=-5.$

Ta có: $f^{\prime }\left(x\right)=\frac{{\left(1-3x+x^2\right)}^{\prime }\left(x-1\right)-\left(1-3x+x^2\right){\left(x-1\right)}^{\prime }}{{\left(x-1\right)}^2}$

$=\frac{\left(-3+2x\right)\left(x-1\right)-\left(1-3x+x^2\right)}{{\left(x-1\right)}^2}=\frac{x^2-2x+2}{{\left(x-1\right)}^2}$.

Bất phương trình $f^{\prime }\left(x\right)>0\iff \frac{x^2-2x+2}{{\left(x-1\right)}^2}>0\iff \{\begin{array}{l}{x}^2-2x+2>0\\ x\ne 1\end{array}\iff x\in ℝ\backslash \left\{1\right\}$.

Ta có $f^{\prime }\left(x\right)=\frac{{\left(x^3\right)}^{\prime }\left(x-1\right)-x^3{\left(x-1\right)}^{\prime }}{{\left(x-1\right)}^2}=\frac{3x^2\left(x-1\right)-x^3}{{\left(x-1\right)}^2}=\frac{2x^3-3x^2}{{\left(x-1\right)}^2}$.

Phương trình $f^{\prime }\left(x\right)=0\iff \frac{2x^3-3x^2}{{\left(x-1\right)}^2}=0\iff 2x^3-3x^2=0\iff [\begin{array}{l}x=0\\ x=\frac{3}{2}\end{array}$.

Ta có $y^{\prime }=\frac{-{\left(x^2-2x+5\right)}^{\prime }}{{\left(x^2-2x+5\right)}^2}=\frac{-2x+2}{{\left(x^2-2x+5\right)}^2}$.

Kiểm tra đáp án A: y=\frac{x^3-1}{x}=x^2-\frac{1}{x}\Rightarrow y^{\prime }=2x+\frac{1}{x^2} (đúng)

Ta có $y^{\prime }=\frac{{\left(2x+5\right)}^{\prime }\left(x^2+3x+3\right)-\left(2x+5\right){\left(x^2+3x+3\right)}^{\prime }}{{\left(x^2+3x+3\right)}^2}$

$=\frac{2\left(x^2+3x+3\right)-\left(2x+5\right)\left(2x+3\right)}{{\left(x^2+3x+3\right)}^2}=\frac{-2x^2-10x-9}{{\left(x^2+3x+3\right)}^2}$.

Ta có: $y^{\prime }=\frac{{\left(-2x^2+x-7\right)}^{\prime }\left(x^2+3\right)-{\left(x^2+3\right)}^{\prime }\left(-2x^2+x-7\right)}{{\left(x^2+3\right)}^2}$

$y^{\prime }=\frac{\left(-4x+1\right)(x^2+3)-2x.\left(-2x^2+x-7\right)}{{\left(x^2+3\right)}^2}=\frac{-x^2+2x+3}{{(x^2+3)}^2}$

Ta có $y=-2\sqrt{x}+3x\Rightarrow y\text{'}=\frac{-1}{\sqrt{x}}+3.$

Do đó $y\text{'}>0\iff \frac{-1}{\sqrt{x}}+3>0\iff 3>\frac{1}{\sqrt{x}}\iff x>\frac{1}{9}$

Ta có f\text{'}\left(x\right)=\frac{1}{2\sqrt{x-1}}.

Tại $x=1$ thì $f\text{'}\left(x\right)$ không xác định.