f(x)= $sin^{6}x+cos^6x$ g(x) $3six^{2}x.cos^2x$ tính f'(x)+g'(x)

0 bình luận về “f(x)= $sin^{6}x+cos^6x$ g(x) $3six^{2}x.cos^2x$ tính f'(x)+g'(x)”

1. $\displaystyle \begin{array}{{>{\displaystyle}l}} f'( x) =6sin^{5} xcosx-6cos^{5} xsinx\\ g'( x) =\left[ 3( sinxcosx)^{2}\right] ‘=\left[\frac{3}{4}( sin2x)^{2}\right] ‘\\ =\frac{3}{4} .2sin2x.2cos2x=3sin2xcos2x\\ f'( x) +g'( x) =6sin^{5} xcosx-6cos^{5} xsinx+3sin2xcos2x\\ =3sin^{4} xsin2x-3cos^{4} xsin2x+3sin2xcos2x\\ =3sin2x\left( sin^{4} x-cos^{4} x\right) +3sin2xcos2x\\ =3sin2x\left( sin^{4} x-cos^{4} x+cos2x\right)\\ =3sin2x\left( sin^{2} x-cos^{2} x+cos^{2} x-sin^{2} x\right)\\ =3sin2x.0=0 \end{array}$

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2. $f(x)=\sin^6x+\cos^6x$

   $=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$

   $=(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x$

   $=1-3\sin^2x\cos^2x$

   $=1-\dfrac{3}{4}.\sin^22x$

   $=1-\dfrac{3}{4}.\dfrac{1-\cos4x}{2}$

   $=\dfrac{3}{8}\cos4x+\dfrac{5}{8}$

   $\to f'(x)=-\dfrac{3}{8}.\sin4x.(4x)’=\dfrac{-3}{2}\sin4x$

   $g(x)=3\sin^2x\cos^2x$

   $=\dfrac{3}{4}.\sin^22x$

   $=\dfrac{3}{4}.\dfrac{1-\cos4x}{2}$

   $=-\dfrac{3}{8}\cos4x+\dfrac{3}{8}$

   $\to g'(x)=\dfrac{3}{8}.\sin4x.(4x)’=\dfrac{3}{2}\sin4x$

   Vậy $f'(x)+g'(x)=0$

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