tính đạo hàm
a) \(y=-3cosx\)
b) \(y=6sinx-5cosx\)
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PL
1
4 tháng 8 2019
a) Ta có : -\(\sqrt{a^2+b^2}< =asinx+bcosx< =\sqrt{a^2+b^2}\)
=> \(-\sqrt{12^2+\left(-5\right)^2}< =y< =\sqrt{12^2+\left(-5\right)^2}\)
<=> \(-\sqrt{13}< =y< =\sqrt{13}\)
Vậy min=\(-\sqrt{13}\) ,max=\(\sqrt{13}\)
b) \(-\sqrt{9+16}< =3cosx-4sinx< =\sqrt{9+16}\)
<=> -5 <=3cos x -4sinx <= 5
<=> 0<= y <= 10
Vậy min=0 max=10
13 tháng 11 2023
a: ĐKXĐ: \(\left(x+2\right)\left(x+3\right)>=0\)
=>\(\left[{}\begin{matrix}x>=-2\\x< =-3\end{matrix}\right.\)
\(y=\sqrt{\left(x+2\right)\left(x+3\right)}=\sqrt{x^2+5x+6}\)
=>\(y'=\dfrac{\left(x^2+5x+6\right)'}{2\sqrt{x^2+5x+6}}=\dfrac{2x+5}{2\sqrt{x^2+5x+6}}\)
y'>0
=>\(\dfrac{2x+5}{2\sqrt{x^2+5x+6}}>0\)
=>2x+5>0
=>\(x>-\dfrac{5}{2}\)
Kết hợp ĐKXĐ, ta được: x>=-2
Đặt y'<0
=>2x+5<0
=>2x<-5
=>\(x< -\dfrac{5}{2}\)
Kết hợp ĐKXĐ, ta được: x<=-3
Vậy: Hàm số đồng biến trên \([-2;+\infty)\) và nghịch biến trên \((-\infty;-3]\)
b: ĐKXĐ: \(\dfrac{2x+1}{x-3}>=0\)
=>\(\left[{}\begin{matrix}x>3\\x< =-\dfrac{1}{2}\end{matrix}\right.\)
\(y=\sqrt{\dfrac{2x+1}{x-3}}\)
=>\(y'=\dfrac{\left(\dfrac{2x+1}{x-3}\right)'}{2\sqrt{\dfrac{2x+1}{x-3}}}\)
=>\(y'=\dfrac{\dfrac{\left(2x+1\right)'\left(x-3\right)-\left(2x+1\right)\left(x-3\right)'}{\left(x-3\right)^2}}{2\sqrt{\dfrac{2x+1}{x-3}}}\)
=>\(y'=\dfrac{\dfrac{2\left(x-3\right)-2x-1}{\left(x-3\right)^2}}{2\sqrt{\dfrac{2x+1}{x-3}}}\)
\(=-\dfrac{\dfrac{7}{\left(x-3\right)^2}}{2\sqrt{\dfrac{2x+1}{x-3}}}< 0\forall x\) thỏa mãn ĐKXĐ, trừ x=-1/2 ra
=>Hàm số luôn đồng biến trên \(\left(3;+\infty\right);\left(-\infty;-\dfrac{1}{2}\right)\)
c:
ĐKXĐ: x>=-3
\(y=\left(x+1\right)\sqrt{x+3}\)
=>\(y'=\left(x+1\right)'\cdot\sqrt{x+3}+\left(x+1\right)\cdot\sqrt{x+3}'\)
=>\(y'=\sqrt{x+3}+\left(x+1\right)\cdot\dfrac{\left(x+3\right)'}{2\sqrt{x+3}}\)
=>\(y'=\sqrt{x+3}+\dfrac{x+1}{2\sqrt{x+3}}\)
=>\(y'=\dfrac{2x+6+x+1}{2\sqrt{x+3}}=\dfrac{3x+7}{2\sqrt{x+3}}\)
Đặt y'>0
=>3x+7>0
=>x>-7/3
Kết hợp ĐKXĐ, ta được: x>-7/3
Đặt y'<0
3x+7<0
=>x<-7/3
Kết hợp ĐKXĐ, ta được: \(-3< x< -\dfrac{7}{3}\)
Vậy: Hàm số đồng biến trên \(\left(-\dfrac{7}{3};+\infty\right)\) và nghịch biến trên \(\left(-3;-\dfrac{7}{3}\right)\)
d: \(y=\dfrac{x-1}{x^2+1}\)(ĐKXĐ: \(x\in R\))
=>\(y'=\dfrac{\left(x-1\right)'\left(x^2+1\right)-\left(x-1\right)\left(x^2+1\right)'}{\left(x^2+1\right)^2}\)
=>\(y'=\dfrac{x^2+1-2x\left(x-1\right)}{\left(x^2+1\right)^2}=\dfrac{-x^2+2x+1}{\left(x^2+1\right)^2}\)
Đặt y'>0
=>\(-x^2+2x+1>0\)
=>\(1-\sqrt{2}< x< 1+\sqrt{2}\)
Đặt y'<0
=>\(-x^2+2x-1< 0\)
=>\(\left[{}\begin{matrix}x>1+\sqrt{2}\\x< 1-\sqrt{2}\end{matrix}\right.\)
Vậy: hàm số đồng biến trên khoảng \(\left(1-\sqrt{2};1+\sqrt{2}\right)\)
hàm số nghịch biến trên khoảng \(\left(1+\sqrt{2};+\infty\right);\left(-\infty;1-\sqrt{2}\right)\)
12 tháng 11 2023
a: \(y=\left(x-1\right)^3\)
=>\(y'=\left[\left(x-1\right)^3\right]'=3\left(x-1\right)^2\cdot\left(x-1\right)'\)
\(=3\left(x-1\right)^2\)
b: \(y=\left(x+2\right)\left(2x^2-3\right)\)
=>\(y'=\left(x+2\right)'\left(2x^2-3\right)+\left(x+2\right)\left(2x^2-3\right)'\)
=>\(y'=2x^2-3+2\left(x+2\right)\)
\(=2x^2+2x+1\)
c: \(y=\left(x-1\right)^2\left(x+2\right)\)
=>\(y=\left(x^2-2x+1\right)\left(x+2\right)\)
=>\(y'=\left(x^2-2x+1\right)'\left(x+2\right)-\left(x^2-2x+1\right)\left(x+2\right)'\)
=>\(y'=\left(2x-2\right)\left(x+2\right)-x^2+2x-1\)
\(=2x^2+4x-2x-4-x^2+2x-1\)
=>\(y'=x^2+4x-5\)
c: \(y=\left(x^2-1\right)\left(2x+1\right)\)
=>\(y'=\left(x^2-1\right)'\left(2x+1\right)+\left(x^2-1\right)\left(2x+1\right)'\)
\(=2x\left(2x+1\right)+2\left(x^2-1\right)\)
\(=4x^2+2x+2x^2-2=6x^2+2x-2\)
JE
đạo hàm
a) \(y=\sqrt{\dfrac{2x-1}{x+1}}\)
b) \(y=4x+\dfrac{3}{2}x^2\)
c) \(y=\dfrac{x^3}{3}-4x^2+7x+1\)
1
HT
23 tháng 4 2021
a/ \(y'=\dfrac{1}{2}.\sqrt{\dfrac{x+1}{2x-1}}.\left(\dfrac{2x-1}{x+1}\right)'=\dfrac{1}{2}\sqrt{\dfrac{x+1}{2x-1}}.\dfrac{3}{\left(x+1\right)^2}\)
b/ \(y'=4+3x\)
c/ \(y'=x^2-8x+7\)
12 tháng 11 2023
a: \(y=\left(x+2\right)\left(2x^2-3\right)\)
=>\(y'=\left(x+2\right)'\left(2x^2-3\right)+\left(x+2\right)\left(2x^2-3\right)'\)
=>\(y'=2x^2-3+\left(x+2\right)\cdot2x\)
\(\Leftrightarrow y'=2x^2-3+2x^2+4x=4x^2+4x-3\)
b: \(y=\left(x-1\right)^2\left(x+2\right)\)
=>\(y=\left(x^2-2x+1\right)\left(x+2\right)\)
=>\(y'=\left(x^2-2x+1\right)'\left(x+2\right)+\left(x^2-2x+1\right)\left(x+2\right)'\)
=>\(y'=\left(2x-2\right)\left(x+2\right)+\left(x^2-2x+1\right)\)
=>\(y'=2x^2+4x-2x-4+x^2-2x+1\)
=>\(y'=3x^2-3\)
c: \(y=\left(x^2-1\right)\left(2x+1\right)\)
=>\(y'=\left(x^2-1\right)'\left(2x+1\right)+\left(x^2-1\right)\left(2x+1\right)'\)
=>\(y'=2x\left(2x+1\right)+2\left(x^2-1\right)\)
=>\(y'=4x^2+2x+2x^2-2=6x^2+2x-2\)
d: \(y=\left(x+2\right)\left(2x^2-5\right)\)
=>\(y'=\left(x+2\right)'\left(2x^2-5\right)+\left(x+2\right)\left(2x^2-5\right)'\)
=>\(y'=2x^2-5+2x\left(x+2\right)=4x^2+4x-5\)
12 tháng 11 2023
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
PT
1
NT
1
NN
1
10 tháng 9 2018
\(A=\dfrac{sinx-3cosx}{2sinx+5cosx}=\dfrac{sinx-3cosx}{sinx}:\dfrac{2sinx+5cosx}{sinx}=\left(1-3.\dfrac{1}{2}\right):\left(2+5.\dfrac{1}{2}\right)=-\dfrac{1}{2}.\dfrac{2}{9}=-\dfrac{1}{9}\)
a: \(y=-3\cdot cosx\)
=>\(y'=-3\cdot\left(-1\right)\cdot sinx=3\cdot sinx\)
b: \(y=6\cdot sinx-5\cdot cosx\)
=>\(y'=6\cdot cosx-5\cdot\left(-1\right)\cdot sinx=6cosx+5\cdot sinx\)