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1/ \(y'=\frac{\sqrt{9-x^2}-x\left(\sqrt{9-x^2}\right)'}{9-x^2}=\frac{\sqrt{9-x^2}+\frac{x^2}{\sqrt{9-x^2}}}{9-x^2}=\frac{9}{\left(9-x^2\right)\sqrt{9-x^2}}\)
2/ \(y'=\frac{\left(\sqrt{x^2+x+3}\right)'.\left(2x+1\right)-2\sqrt{x^2+x+3}}{\left(2x+1\right)^2}=\frac{\frac{\left(2x+1\right)}{2\sqrt{x^2+x+3}}.\left(2x+1\right)-2\sqrt{x^2+x+3}}{\left(2x+1\right)^2}\)
\(=\frac{\left(2x+1\right)^2-4\left(x^2+x+3\right)}{2\left(2x+1\right)^2\sqrt{x^2+x+3}}=\frac{-11}{2\left(2x+1\right)^2\sqrt{x^2+x+3}}\)
3/ \(y'=3\left(1+tan^23x\right)=3+3tan^23x\)
4/ \(y'=\frac{\left(cosx-sinx\right)\left(sinx-cosx\right)-\left(cosx+sinx\right)\left(sinx+cosx\right)}{\left(sinx-cosx\right)^2}\)
\(=-\frac{\left(sinx-cosx\right)^2+\left(sinx+cosx\right)^2}{\left(sinx-cosx\right)^2}=-\frac{sin^2x+cos^2x-2sinxcosx+sin^2x+cos^2x+2sinxcosx}{sin^2x+cos^2x-2sinxcosx}\)
\(=\frac{-2}{1-sin2x}\)
5/ \(y'=4x+\frac{1}{2\sqrt{x}}-\frac{\pi}{2}cos\left(\frac{\pi x}{2}\right)\)
6/ \(y'=3sin^2\left(1-3x\right).\left(sin\left(1-3x\right)\right)'=3sin^2\left(1-3x\right).cos\left(1-3x\right).\left(1-3x\right)'\)
\(=-9sin^2\left(1-3x\right).cos\left(1-3x\right)\)
Lời giải (Giao lưu_cách làm cấp 2)
\(f'\left(x\right)=6x^8-6x^5+6x^2-6x+6=6\left(x^8-x^5+x^2-x+1\right)=6A\)
Cần c/m : \(A>\left(x^8-x^5+x^2-x+1\right)...với\forall x\in R\)
Nếu \(\left|x\right|\ge1\Rightarrow\left\{{}\begin{matrix}x^8\ge x^5\\x^2\ge x\end{matrix}\right.\) \(\Rightarrow A=\left(x^8-x^5\right)+\left(x^2-x\right)+1>0\Rightarrow A>0\)(1)
Nếu \(\left|x\right|< 1\Rightarrow\left\{{}\begin{matrix}x^2>x^5\\1>x\end{matrix}\right.\)\(\Rightarrow A=\left(x^2-x^5\right)+\left(1-x\right)+x^8>0\Rightarrow A>0\)(2)
Từ (1) và (2) \(\Rightarrow A>0\forall x\in R\)=> dpcm
Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra f'(x)=0
a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0
d,f(x)=\(\frac{3}{2}\)=>f'(x)=0
\(f'=6x^8-6x^5+6x+6=6\left(x^8-x^5+x+1\right)\)
\(\left[{}\begin{matrix}\left|x\right|\le1\Rightarrow\left|x^5-x\right|\le\left|x\right|\le1\Rightarrow1-x^5-x\ge0\\\left|x\right|\ge1\Rightarrow\left|x^5\right|\le x^8\Rightarrow\left\{{}\begin{matrix}x^8-x^5>0\\x^2-x>0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow f'\left(x\right)>0\forall x\)
a: \(-1< =cosx< =1\)
\(\Leftrightarrow-2< =2cosx< =2\)
\(\Leftrightarrow-5< =2cosx-3< =-1\)
\(f\left(x\right)_{min}=-5\) khi cos x=-1
hay \(x=\Pi+k2\Pi\)
\(f\left(x\right)_{max}=-1\) khi cos x=1
hay \(x=k2\Pi\)
b: \(-1< =sinx< =1\)
\(\Leftrightarrow-2< =2sinx< =2\)
\(\Leftrightarrow5< =2sinx+7< =9\)
\(\Leftrightarrow\sqrt{5}< =\sqrt{2sinx+7}< =3\)
\(\Leftrightarrow3\sqrt{5}< =3\sqrt{2sinx+7}< =9\)
\(f\left(x\right)_{min}=3\sqrt{5}\) khi sin x=-1
hay \(x=-\dfrac{\Pi}{2}+k2\Pi\)
\(f\left(x\right)_{max}=9\) khi sin x=1
hay \(x=\dfrac{\Pi}{2}+k2\Pi\)
a) f'(x) = - 3sinx + 4cosx + 5. Do đó
f'(x) = 0 <=> - 3sinx + 4cosx + 5 = 0 <=> 3sinx - 4cosx = 5
<=> sinx -
cosx = 1. (1)
Đặt cos φ = , (φ ∈
) => sin φ =
, ta có:
(1) <=> sinx.cos φ - cosx.sin φ = 1 <=> sin(x - φ) = 1
<=> x - φ = + k2π <=> x = φ +
+ k2π, k ∈ Z.
b) f'(x) = - cos(π + x) - sin = cosx + sin
.
f'(x) = 0 <=> cosx + sin = 0 <=> sin
= - cosx <=> sin
= sin
<=> =
+ k2π hoặc
= π - x +
+ k2π
<=> x = π - k4π hoặc x = π + k, (k ∈ Z).
f'(1) = -8
f'(2) = 0
f'(3) = 0