cho hso \(y=x^3-3x^2+3x+1\). Đạo hàm của hàm số đạt gtnn khi??
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HT
23 tháng 4 2021
\(y'=-3x^2-6x+m\Rightarrow y''=-6x-6\)
\(y''=0\Leftrightarrow-6x-6=0\Leftrightarrow x=-1\notin\left[0;1\right]\)
\(\left\{{}\begin{matrix}y'\left(0\right)=m\\y'\left(1\right)=m-9\end{matrix}\right.\Rightarrow^{max}_{\left[0;1\right]}y'=y'\left(0\right)=m\)
\(\Rightarrow m=10\)
31 tháng 8 2023
a: \(y'=\left(x^2\right)'+\left(3x\right)'-\left(6x^6\right)'+\left(\dfrac{2x-3}{x-1}\right)'\)
\(=2x+3-6\cdot6x^5+\dfrac{\left(2x-3\right)'\left(x-1\right)-\left(2x-3\right)\left(x-1\right)'}{\left(x-1\right)^2}\)
\(=-36x^5+2x+3+\dfrac{2\left(x-1\right)-2x+3}{\left(x-1\right)^2}\)
\(=-36x^5+2x+3+\dfrac{1}{\left(x-1\right)^2}\)
b: \(\left(\sqrt{2x^2-3x+1}\right)'=\dfrac{\left(2x^2-3x+1\right)'}{2\sqrt{2x^2-3x+1}}\)
\(=\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)
\(y'=3\cdot2x-4+\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)
\(=6x-4+\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)
c: \(\left(\sqrt{4x^2-3x+1}\right)'=\dfrac{\left(4x^2-3x+1\right)'}{2\sqrt{4x^2-3x+1}}\)
\(=\dfrac{8x-3}{2\sqrt{4x^2-3x+1}}\)
\(y'=\left(\sqrt{4x^2-3x+1}\right)'-4'=\dfrac{8x-3}{2\sqrt{4x^2-3x+1}}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2022
\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)
\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)
\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)
\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
20 tháng 8 2023
a: \(y'=\left(x^2+3x-1\right)'\cdot e^x+\left(x^2+3x-1\right)\cdot\left(e^x\right)'\)
\(=e^x\left(2x+3\right)+\left(x^2+3x-1\right)\cdot e^x\)
\(=e^x\left(x^2+5x+2\right)\)
b: \(y'=\left(x^3\right)'\cdot log_2x+x^3\cdot\left(log_2x\right)'\)
\(=3x^2\cdot log_2x+x^3\cdot\dfrac{1}{x\cdot ln2}\)
\(y'=3x^2-6x+3=3\left(x^2-2x+1\right)=3\left(x-1\right)^2\ge0\)
\("="\Leftrightarrow x=1\)