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

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

3 - \tan x

Applying the product rule for logarithms converts

y = \ln(e^{3x} \cos x)

into

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

. Differentiating

3x

yields

3

, and differentiating

\ln(\cos x)

by chain rule yields

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

. Combining these gives

3 - \tan x

Simplify the logarithmic function using log laws

y = \ln(e^{3x}) + \ln(\cos x) = 3x + \ln(\cos x)

Splitting the logarithm of a product simplifies the differentiation process.

Differentiate each term with respect to

x

\frac{dy}{dx} = \frac{d}{dx}(3x) + \frac{d}{dx}(\ln(\cos x)) = 3 + \frac{1}{\cos x} \cdot (-\sin x)

The derivative of

3x

3

, and applying the chain rule to

\ln(u)

gives

\frac{1}{u} \cdot \frac{du}{dx}

where

u = \cos x

Simplify the trigonometric ratio

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

Replacing

\frac{\sin x}{\cos x}

with

\tan x

yields the simplified final derivative.

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule and Log Laws

Question