Question

solve this question
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Answer 1

Step-by-step explanation:

$y = (cos) {}^{y} \\ taking \: log \: both \: sides \: log \: (y) = ylog(cosx) \\ differentiating \: with \: respect \: x \: \\ \frac{1dy}{y \: dx} = y \frac{( - sinx)}{cosx} + \frac{dy}{dx} log(cosx) \\ \frac{1 \: dy}{y \: dx} (1 - ylog(cosx)) = \frac{ - ysinx}{cosx} = - ytanx \\ or \frac{dy}{dx } = \frac{ - y {}^{2}tanx }{1 - ylog(cosx)}$