f(x)=sinx^4+cosx^4
g(x)=1/4.cos4x
CMR: f'(x)=g'(x)
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AH
Akai Haruma
Giáo viên
11 tháng 3 2018
Lời giải:
Ta có: \(f(x)=\sin ^4x+\cos ^4x=(\sin ^2x)^2+(\cos ^2x)^2+2\sin ^2x\cos ^2x-2\sin ^2x\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)^2-\frac{1}{2}(2\sin x\cos x)^2\)
\(=1-\frac{1}{2}\sin ^2(2x)\)
Do đó: \(f'(x)=[1-\frac{1}{2}\sin ^2(2x)]'=-\frac{1}{2}.2.\sin 2x(\sin 2x)'\)
\(=-2\sin 2x.\cos 2x=-\sin 4x\)
Và: \(g(x)=\frac{1}{4}(\cos 4x)\Rightarrow g'(x)=\frac{1}{4}.(4x)'-\sin (4x)=-\sin 4x\)
Do đó: \(f'(x)=g'(x)\)
PN
1
KR
28 tháng 1 2023
\(I= \int \frac{sinx-cosx}{(sinx+cosx)^2-4}\ dx \\u=sinx+cosx, du=(cosx-sinx) dx=-(sinx-cosx)dx \\I = -\int \frac{du}{u^2-4} \\ =-\int \frac{\frac{1}{4}}{u-2}+\frac{\frac{1}{4}}{u+2}\ du \\ = -\frac{1}{4}ln(|\frac{sinx+cosx-2}{sinx+cosx+2}|)+C\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 3 2019
\(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{sin^2x.cosx+2sin2x}{\left(f\left(sinx\right)\right)^2}dx=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\left(sin^2x+4sinx\right).cosx}{\left(f\left(sinx\right)\right)^2}dx\)
Đặt \(sinx=t\Rightarrow cosx.dx=dt;\left\{{}\begin{matrix}x=\frac{\pi}{6}\Rightarrow t=\frac{1}{2}\\x=\frac{\pi}{3}\Rightarrow t=\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\frac{\left(t^2+4t\right)}{f^2\left(t\right)}dt=\int\limits^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\frac{\left(x^2+4x\right)}{f^2\left(x\right)}dx\)
Lại có:
\(x+x.f'\left(x\right)=2f\left(x\right)-4\Leftrightarrow x+4=2f\left(x\right)-x.f'\left(x\right)\)
\(\Leftrightarrow x^2+4x=2x.f\left(x\right)-x^2.f'\left(x\right)\)
\(\Leftrightarrow\frac{x^2+4x}{f^2\left(x\right)}=\frac{2x.f\left(x\right)-x^2.f'\left(x\right)}{f^2\left(x\right)}=\left(\frac{x^2}{f\left(x\right)}\right)'\)
\(\Rightarrow I=\int\limits^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\left(\frac{x^2}{f\left(x\right)}\right)'dx=\frac{x^2}{f\left(x\right)}|^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}=\frac{\left(\frac{\sqrt{3}}{2}\right)^2}{f\left(\frac{\sqrt{3}}{2}\right)}-\frac{\left(\frac{1}{2}\right)^2}{f\left(\frac{1}{2}\right)}=\frac{3}{4b}-\frac{1}{4a}\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2019
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/ \(y'=3sin^2\left(1-3x\right).\left(sin\left(1-3x\right)\right)'=3sin^2\left(1-3x\right).cos\left(1-3x\right).\left(1-3x\right)'\)
\(=-9sin^2\left(1-3x\right).cos\left(1-3x\right)\)