3 years ago

How do you find the derivative of y=ln(cos(x))y=ln(cos(x)) ?

1 answer:

7 0

$y=ln(cosx)\\\\We\ know:\\(lnx)'=\frac{1}{x}\ and\ (cosx)'=-sinx\ and\ \{f[g(x)]\}'=f'[g(x)]\cdot g'(x)\\\\therefore:\\\\y'=\frac{1}{cosx}\cdot(-sinx)=-\frac{sinx}{cosx}=-tanx$

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IgorC [24]

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<h2>y + 3 = -6(x + 9) - point-slope form</h2><h2>y = -6x - 57 - slope-intercept form</h2>

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========================================

We have the slope <em>m = -6</em> and the point (-9, -3).

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