Câu 13.
\(y'=\left[sin\left(\dfrac{\pi}{6}-3x\right)\right]'=\left(\dfrac{\pi}{6}-3x\right)'.cos\left(\dfrac{\pi}{6}-3x\right)=-3cos\left(\dfrac{\pi}{6}-3x\right)\)
Câu 14.
\(y'=\left[-\dfrac{1}{2}sin\left(\dfrac{\pi}{3}-x^2\right)\right]'=-\dfrac{1}{2}\left(\dfrac{\pi}{3}-x^2\right)'.cos\left(\dfrac{\pi}{3}-x^2\right)\\ =-\dfrac{1}{2}.\left(-2x\right).cos\left(\dfrac{\pi}{3}-x^2\right)=x.cos\left(\dfrac{\pi}{3}-x^2\right)\)
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