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a: \(y'=4\cdot3x^2-3\cdot2x+2=12x^2-6x+2\)
b: \(y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}=\dfrac{x-1-x-1}{\left(x-1\right)^2}=\dfrac{-2}{\left(x-1\right)^2}\)
c: \(y'=-2\cdot\left(\sqrt{x}\cdot x\right)'\)
\(=-2\cdot\left(\dfrac{x+x}{2\sqrt{x}}\right)=-2\cdot\dfrac{2x}{2\sqrt{x}}=-2\sqrt{x}\)
d: \(y'=\left(3sinx+4cosx-tanx\right)\)'
\(=3cosx-4sinx+\dfrac{1}{cos^2x}\)
e: \(y'=\left(4^x+2e^x\right)'\)
\(=4^x\cdot ln4+2\cdot e^x\)
f: \(y'=\left(x\cdot lnx\right)'=lnx+1\)
\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)
\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)
3.
\(f\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{\pi}{3}\right)\Rightarrow f'\left(x+\frac{\pi}{3}\right)=-sin\left(x+\frac{\pi}{3}\right)\)
\(f'\left(x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{6}\right)\)
\(f'\left(0\right)=-sin\left(0\right)=0\)
\(2f'\left(x+\frac{\pi}{3}\right).f'\left(x-\frac{\pi}{6}\right)=2sin\left(x+\frac{\pi}{3}\right)sin\left(x-\frac{\pi}{6}\right)\)
\(=cos\left(\frac{\pi}{2}\right)-cos\left(2x+\frac{\pi}{6}\right)=-cos\left(2x+\frac{\pi}{6}\right)\)
\(f'\left(0\right)-f\left(2x+\frac{\pi}{6}\right)=0-cos\left(2x+\frac{\pi}{6}\right)=-cos\left(2x+\frac{\pi}{6}\right)\)
\(\Rightarrow2f'\left(x+\frac{\pi}{3}\right)f'\left(x-\frac{\pi}{6}\right)=f'\left(0\right)-f\left(2x+\frac{\pi}{6}\right)\) (đpcm)
4.
\(y=3\left(sin^4x+cos^4x\right)-2\left(sin^6x+cos^6x\right)\)
\(=3\left(sin^2x+cos^2x\right)^2-6sin^2x.cos^2x-2\left(sin^2x+cos^2x\right)^3+6sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(=3-2=1\)
\(\Rightarrow y'=0\) ; \(\forall x\)
5.
\(y=\left(\frac{sinx}{1+cosx}\right)^3=\left(\frac{sinx\left(1-cosx\right)}{1-cos^2x}\right)^3=\left(\frac{sinx\left(1-cosx\right)}{sin^2x}\right)^3=\left(\frac{1-cosx}{sinx}\right)^3\)
\(y'=3\left(\frac{1-cosx}{sinx}\right)^2\left(\frac{sin^2x-cosx\left(1-cosx\right)}{sin^2x}\right)=3\left(\frac{1-cosx}{sinx}\right)^2\left(\frac{1-cosx}{sin^2x}\right)=\frac{3\left(1-cosx\right)^3}{sin^4x}\)
\(\Rightarrow y'.sinx-3y=\frac{3\left(1-cosx\right)^3}{sin^3x}-3\left(\frac{1-cosx}{sinx}\right)^3=0\) (đpcm)
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: \(y=ln^2x\)
=>\(y'=\dfrac{\left(-1\right)}{x^2}=-\dfrac{1}{x^2}\)
Chọn C
f ' x = − 2 cos x sin x 1 + sin 2 x − 2 cos x sin x cos 2 x 1 + sin 2 x 2
= − 2 cos x sin x 1 + sin 2 x + cos 2 x 1 + sin 2 x 2 = − 4 cos x sin x 1 + sin 2 x 2
⇒ f ' π 4 = − 8 9