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I understand how the product rule works, but I do not know what to do when 'ln' comes into it. If somebody could explain how to do this and what rules are involved I would really appreciate it.

Find derivative of: $\ln(\cos(x))$

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    This has nothing to deal with the product rule. You need the chain rule. See https://en.wikipedia.org/wiki/Chain_rule – 2017-02-27
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    Hint. The chain rule. – 2017-02-27
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    Use $(\ln u)'=\dfrac{u'}{u}$. – 2017-02-27

4 Answers 4

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the derivative of $$\ln(\cos(x))$$ is given by $$-\frac{\sin(x)}{\cos(x)}$$ after the chain rule and the derivative of $\ln(x)$ is equal to $$\frac{1}{x}$$

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Use the chain rule $(\ln(u))'=\dfrac{u'}{u}$. The you get $(\ln(\cos(x))'=\dfrac{-\sin(x)}{\cos(x)}=-\tan(x) $.

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    It should be $-\tan(x)$ – 2017-02-27
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    Of course. I apologize. – 2017-02-27
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You need to use the chain rule which states:

$$ (f \circ g)'(x) = f'(g(x)) \cdot g'(x)$$

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Well, we can use the chain rule:

$$\frac{\text{d}}{\text{d}x}\left(\text{f}\left(\text{g}\left(x\right)\right)\right)=\text{g}'\left(x\right)\cdot\text{f}\space'\left(\text{g}\left(x\right)\right)\tag1$$

So, we get for your problem:

$$\frac{\text{d}}{\text{d}x}\left(\ln\left(\cos\left(x\right)\right)\right)=\frac{\text{d}}{\text{d}x}\left(\cos\left(x\right)\right)\cdot\frac{1}{\cos\left(x\right)}=-\sin\left(x\right)\cdot\sec\left(x\right)=-\tan\left(x\right)\tag2$$