satyamshah365 satyamshah365

12-10-2021
Mathematics

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e^3x.cos4x.sin2x hii please solv this you have to differentiate

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Аноним Аноним

12-10-2021

Answer:

[tex]y = e^{3x}.cos4x.sin2x[/tex]

[tex]=> y' =- 3e^{3x}.4sin4x.2cos2x=-24.e^{3x}.sin4x.cos2x[/tex]

[tex]dy=y'.dx => dy = -24e^{3x}.sin4x.cos2x.dx[/tex]

Step-by-step explanation:

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