Hộ mình tìm GTLN,GTLN của hàm số sau
\(1.y=\sqrt{5-2cos^2xsin^2x}\)
\(2.y=cos^2x+2cos2x\)
\(3.y=sin^2x-4sinx+3\)
\(4.y=2\sqrt{s\text{inx}}+3\)
\(5.y=\sqrt{1+c\text{os}\left(4x\right)^2}-2\)
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1.
\(y=\sqrt{5-2\cos ^2x\sin ^2x}=\sqrt{5-\frac{1}{2}(2\cos x\sin x)^2}=\sqrt{5-\frac{1}{2}\sin ^22x}\)
Dễ thấy:
$\sin ^22x\geq 0\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\leq \sqrt{5}$
Vậy $y_{\max}=\sqrt{5}$
$\sin ^22x\leq 1\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\geq \sqrt{5-\frac{1}{2}}=\frac{3\sqrt{2}}{2}$
Vậy $y_{\min}=\frac{3\sqrt{2}}{2}$
2.
$y=1+\frac{1}{2}\sin 2x\cos 2x=1+\frac{1}{4}.2\sin 2x\cos 2x$
$=1+\frac{1}{4}\sin 4x$
Vì $-1\leq \sin 4x\leq 1$
$\Rightarrow \frac{5}{4}\leq 1+\frac{1}{4}\sin 4x\leq \frac{3}{4}$
$\Leftrightarrow \frac{5}{4}\leq y\leq \frac{3}{4}$
Vậy $y_{\max}=\frac{5}{4}; y_{\min}=\frac{3}{4}$
Coi như tất cả các biểu thức cần tính đạo hàm đều xác định.
1.
\(y'=2sin\sqrt{4x+3}.\left(sin\sqrt{4x+3}\right)'=2sin\sqrt{4x+3}.cos\sqrt{4x+3}.\left(\sqrt{4x+3}\right)'\)
\(=sin\left(2\sqrt{4x+3}\right).\dfrac{4}{2\sqrt{4x+3}}=\dfrac{2sin\left(2\sqrt{4x+3}\right)}{\sqrt{4x+3}}\)
2.
\(y'=3x^3+\dfrac{17}{x\sqrt{x}}\)
3.
\(y'=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\left(\dfrac{sin4x}{cos\left(x^2+2\right)}\right)'\)
\(=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\dfrac{4cos4x.cos\left(x^2+2\right)+2x.sin4x.sin\left(x^2+2\right)}{cos^2\left(x^2+2\right)}\)
4.
\(y'=-\dfrac{\left(\sqrt{sin^2\left(6-x\right)+4x}\right)'}{sin^2\left(6-x\right)+4x}=-\dfrac{\left[sin^2\left(6-x\right)+4x\right]'}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
\(=-\dfrac{2sin\left(6-x\right).\left[sin\left(6-x\right)\right]'+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}=-\dfrac{-2sin\left(6-x\right).cos\left(6-x\right)+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
\(=\dfrac{sin\left(12-2x\right)-4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
5.
\(y'=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).\left[sin\left(\dfrac{2x-1}{4-x}\right)\right]'\)
\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).cos\left(\dfrac{2x-1}{4-x}\right).\left(\dfrac{2x-1}{4-x}\right)'\)
\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+x.sin\left(\dfrac{4x-2}{4-x}\right).\dfrac{7}{\left(4-x\right)^2}\)
1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)
\(y=2-\left(-cosx\right).\left(-sinx\right)\)
y = 2 - sinx.cosx
y = \(2-\dfrac{1}{2}sin2x\)
Max = 2 + \(\dfrac{1}{2}\) = 2,5
Min = \(2-\dfrac{1}{2}\) = 1,5
2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)
Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)
Max = \(\sqrt{5}\)
a: \(y'=\left(x^2+2x\right)'\left(x^3-3x\right)+\left(x^2+2x\right)\left(x^3-3x\right)'\)
\(=\left(2x+2\right)\left(x^3-3x\right)+\left(x^2+2x\right)\left(3x^2-3\right)\)
\(=2x^4-6x^2+2x^3-6x+3x^4-3x^2+6x^3-6x\)
\(=5x^4+8x^3-9x^2-12x\)
b: y=1/-2x+5
=>\(y'=\dfrac{2}{\left(2x+5\right)^2}\)
c: \(y'=\dfrac{\left(4x+5\right)'}{2\sqrt{4x+5}}=\dfrac{4}{2\sqrt{4x+5}}=\dfrac{2}{\sqrt{4x+5}}\)
d: \(y'=\left(sinx\right)'\cdot cosx+\left(sinx\right)\cdot\left(cosx\right)'\)
\(=cos^2x-sin^2x=cos2x\)
e: \(y=x\cdot e^x\)
=>\(y'=e^x+x\cdot e^x\)
f: \(y=ln^2x\)
=>\(y'=\dfrac{\left(-1\right)}{x^2}=-\dfrac{1}{x^2}\)
a.
\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)
b.
\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)
c.
\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)
d.
\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)
e.
\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)
g.
\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)
3.
\(y=\left(3-sinx\right)\left(1-sinx\right)\ge0\)
\(\Rightarrow y_{min}=0\) khi \(sinx=1\)
\(y=sin^2x-4sinx-5+8=\left(sinx+1\right)\left(sinx-5\right)+8\le8\)
\(y_{max}=8\) khi \(sinx=-1\)
4.
\(0\le\sqrt{sinx}\le1\Rightarrow3\le y\le5\)
\(y_{min}=3\) khi \(sinx=0\)
\(y_{max}=5\) khi \(sinx=1\)
5.
Đề là \(cos^24x\) hay \(cos\left(\left(4x\right)^2\right)\)
Hai biểu thức này cho 2 kết quả khác nhau
1.
\(y=\sqrt{5-\frac{1}{2}\left(2sinx.cosx\right)^2}=\sqrt{5-\frac{1}{2}sin^22x}\)
Do \(0\le sin^22x\le1\) \(\Rightarrow\frac{3\sqrt{2}}{2}\le y\le\sqrt{5}\)
\(y_{min}=\frac{3\sqrt{2}}{2}\) khi \(sin^22x=1\)
\(y_{max}=\sqrt{5}\) khi \(sin2x=0\)
2.
\(y=cos^2x+2\left(2cos^2x-1\right)=5cos^2x-2\)
Do \(0\le cos^2x\le1\Rightarrow-2\le y\le3\)
\(y_{min}=-2\) khi \(cosx=0\)
\(y_{max}=3\) khi \(cos^2x=1\)