If y = 3e^{2x} + \ln(\cos x), calculate the value of \frac{dy}{dx} at x = 0. — JAMB UTME

If $y = 3e^{2x} + \ln(\cos x)$ , calculate the value of $\frac{dy}{dx}$ at $x = 0$ .

Answer: 6

The value of the derivative of

y = 3e^{2x} + \ln(\cos x)

x = 0

is 6.

Differentiating each transcendental term individually using the chain rule yields

\frac{dy}{dx} = 6e^{2x} - \tan x

. Substituting

x = 0

gives

6e^0 - \tan 0 = 6 - 0 = 6

Differentiate the exponential term

3e^{2x}

6e^{2x}

Applying the derivative rule for exponential functions

\frac{d}{dx}[a e^{kx}] = a k e^{kx}

Differentiate the logarithmic term

\ln(\cos x)

-\tan x

Applying the chain rule

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

where

u = \cos x

gives

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

Combine terms and evaluate at

x = 0

Substituting

x = 0

into

\frac{dy}{dx} = 6e^{2x} - \tan x

yields

6e^0 - \tan 0 = 6(1) - 0 = 6

Differentiation of Trigonometric, Exponential, and Logarithmic Functions

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