If y = \ln(\cos(3x)), what is \frac{dy}{dx}? — JAMB UTME

If $y = \ln(\cos(3x))$ , what is $\frac{dy}{dx}$ ?

-3\tan(3x)

Differentiating

y = \ln(\cos(3x))

requires using the chain rule twice: first for

\ln(u)

, giving

\frac{1}{\cos(3x)}

, and then for

\cos(3x)

, giving

-3\sin(3x)

. Multiplying these results yields

\frac{-3\sin(3x)}{\cos(3x)}

, which simplifies to

-3\tan(3x)

Apply the chain rule formula for differentiating logarithmic functions

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

\frac{dy}{dx} = \frac{1}{\cos(3x)} \cdot \frac{d}{dx}(\cos(3x))

The outer function is

\ln(u)

where

u = \cos(3x)

Differentiate the inner trigonometric function

\cos(3x)

\frac{d}{dx}(\cos(3x)) = -3\sin(3x)

By the chain rule,

\frac{d}{dx}[\cos(kx)] = -k\sin(kx)

Multiply the terms and simplify using trigonometric identities

\frac{dy}{dx} = \frac{-3\sin(3x)}{\cos(3x)} = -3\tan(3x)

Using the identity

\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)

, the derivative simplifies to

-3\tan(3x)

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule

Estimated Time:1m 30s

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