Question #9043b

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Question #9043b

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Answer 1 · 2016-12-11T22:08:40

Use $ln (\frac{u}{v}) = ln u - ln v$ $ln(u/v) = lnu-lnv$ to simplify the process.

Explanation:

(If $c$ $c$ is a constant, then the whole thing is constant and the derivative is $0$ $0$ )

I will assume that $c$ $c$ is the independent variable. If we are differentiating with respect to some other variable, then we must apply the chain rule and multiply by the derivative of $c$ $c$ .

$f (c) = ln (\frac{1 + sin c}{1 - sin c}) = ln (1 + sin c) - ln (1 - sin c)$ $f(c) = ln((1+sinc)/(1-sinc)) = ln(1+sinc)-ln(1-sinc)$

Now use $\frac{d}{d c} (ln u) = \frac{1}{u} \frac{d u}{d c}$ $d/(dc) (lnu) = 1/u (du)/(dc)$ (chain rule) to get

$f'(c) = 1/(1+sinc) * (cosc) - 1/(1-sinc) (-cosc)$

$= cosc/(1+sinc) + cosc/(1-sinc)$

$= (2cosc)/(1-sin^2c)$

$= 2/cosc = 2sec c$ .

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Question #9043b

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