Cho hàm số f(x)=cos^2x+cos^2(2pi/3+x)+cos^2(2pi/3-x). Tính đạo hàm f'(x)

Trả lời

Có $f^{\text{'}}\left(x\right)={\left(co{s}^{2}x+co{s}^2\left(\frac{2\pi }{3}+x\right)+co{s}^2\left(\frac{2\pi }{3}-x\right)\right)}^{\text{'}}$

$={\left(co{s}^{2}x\right)}^{\text{'}}+{\left(co{s}^2\left(\frac{2\pi }{3}+x\right)\right)}^{\text{'}}+{\left(co{s}^2\left(\frac{2\pi }{3}-x\right)\right)}^{\text{'}}$

$=2\cos x\cdot {\left(cosx\right)}^{\text{'}}+2cos\left(\frac{2\pi }{3}+x\right)\cdot {\left(cos\left(\frac{2\pi }{3}+x\right)\right)}^{\text{'}}+2cos\left(\frac{2\pi }{3}-x\right)\cdot {\left(cos\left(\frac{2\pi }{3}-x\right)\right)}^{\text{'}}$

$=-2\cos x\cdot \sin x-2\cos \left(\frac{2\pi }{3}+x\right)\sin \left(\frac{2\pi }{3}+x\right)+2\cos \left(\frac{2\pi }{3}-x\right)\sin \left(\frac{2\pi }{3}-x\right)$

$=-\sin 2x-\sin \left(\frac{4\pi }{3}+2x\right)+\sin \left(\frac{4\pi }{3}-2x\right)$

$=-\sin 2x-\sin \left(\pi +\frac{\pi }{3}+2x\right)+\sin \left(\pi +\frac{\pi }{3}-2x\right)$

$=-\sin 2x+\sin \left(\frac{\pi }{3}+2x\right)-\sin \left(\frac{\pi }{3}-2x\right)$$=-\sin 2x+2cos\frac{\pi }{3}\sin 2x$$=-\sin 2x+\sin 2x=0$

Vậy f'(x) = 0 với mọi x Î ℝ.