What is the derivative of ln(cos(x))?

The Mathematician — Tue, 27 Sep 2022 13:00:49 +0000

The derivative of

\ln(\cos(x))

-\tan(x)

. We will see here how we can find that.

Solution. Let

h(x) = \ln(\cos(x))

f(u) = \ln(u)

and

g(x) = \cos(x)

. Then we apply the chain rule, i.e.,

\begin{align*} h'(x) = f'(g(x))g'(x). \end{align*}

We do know that

\frac{d}{dx}\cos(x) = -\sin(x)

and that

\frac{d}{du} \ln(u) = \frac{1}{u}

. So we get

\begin{align*} f'(u) = \frac{1}{u} \quad \text{and} \quad g'(x) = -\sin(x). \end{align*}

In conclusion, we get:

\begin{align*} h'(x) &= f'(g(x))g'(x) \\ &= \frac{1}{\cos(x)}\cdot (-\sin(x)) \\ &= \frac{-\sin(x)}{\cos(x)} \\ &= -\tan(x). \end{align*}

So the derivative of

\ln(\cos(x))

-\tan(x)

ln(cos(x)) Archives - Epsilonify