If y = \ln(\cos 2x), what is \frac{dy}{dx}? — JAMB UTME

If $y = \ln(\cos 2x)$ , what is $\frac{dy}{dx}$ ?

-2\tan 2x

Applying the chain rule to

y = \ln(\cos 2x)

yields

\frac{dy}{dx} = \frac{1}{\cos 2x} \cdot (-\sin 2x) \cdot 2 = -2\frac{\sin 2x}{\cos 2x} = -2\tan 2x

Identify the inner and outer functions for the chain rule.

Let

u = \cos 2x

, so

y = \ln u

The given function

y = \ln(\cos 2x)

is a composite transcendental function.

Differentiate

y

with respect to

u

, and

u

with respect to

x

\frac{dy}{du} = \frac{1}{u} = \frac{1}{\cos 2x}

, and

\frac{du}{dx} = -2\sin 2x

The derivative of

\ln u

\frac{1}{u}

and the derivative of

\cos 2x

-2\sin 2x

using the chain rule.

Apply the chain rule

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

and simplify.

\frac{dy}{dx} = \frac{1}{\cos 2x} \cdot (-2\sin 2x) = -2\left(\frac{\sin 2x}{\cos 2x}\right) = -2\tan 2x

Using the trigonometric identity

\frac{\sin 2x}{\cos 2x} = \tan 2x

simplifies the expression into standard form.

Differentiation of composite logarithmic and trigonometric functions using the Chain Rule

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