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Question-118062




Question Number 118062 by andilizhaa last updated on 15/Oct/20
Commented by bobhans last updated on 15/Oct/20
y′=3×6×sin^5 (3x−π)×cos (3x−π)  y′=9sin^4 (3x−π)sin (6x−2π)
$$\mathrm{y}'=\mathrm{3}×\mathrm{6}×\mathrm{sin}\:^{\mathrm{5}} \left(\mathrm{3x}−\pi\right)×\mathrm{cos}\:\left(\mathrm{3x}−\pi\right) \\ $$$$\mathrm{y}'=\mathrm{9sin}\:^{\mathrm{4}} \left(\mathrm{3x}−\pi\right)\mathrm{sin}\:\left(\mathrm{6x}−\mathrm{2}\pi\right) \\ $$
Commented by andilizhaa last updated on 15/Oct/20
3×6  where from?
$$\mathrm{3}×\mathrm{6}\:\:{where}\:{from}? \\ $$

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