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Question-168787




Question Number 168787 by TOTTI last updated on 17/Apr/22
Commented by Oktamboy last updated on 17/Apr/22
lny=ln3×cosx  ((y′)/y)=−ln3×sinx  y′=−ln3×sinx×3^(cosx)
$${lny}={ln}\mathrm{3}×{cosx} \\ $$$$\frac{{y}'}{{y}}=−{ln}\mathrm{3}×{sinx} \\ $$$${y}'=−{ln}\mathrm{3}×{sinx}×\mathrm{3}^{{cosx}} \\ $$

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