If y = \ln(2 + \sin(3x)) + e^{4x}, find the value of \frac{dy}{dx} at x = 0. — JAMB UTME

If $y = \ln(2 + \sin(3x)) + e^{4x}$ , find the value of $\frac{dy}{dx}$ at $x = 0$ .

Answer: 5.5

The value of

\frac{dy}{dx}

x = 0

5.5

Differentiating

\ln(2 + \sin(3x))

by the chain rule gives

\frac{3\cos(3x)}{2 + \sin(3x)}

, and differentiating

e^{4x}

gives

4e^{4x}

. Evaluating

\frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}

x = 0

yields

\frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5

Differentiate the logarithmic component

\ln(2 + \sin(3x))

using the chain rule

\frac{d}{dx}[\ln(2 + \sin(3x))] = \frac{3\cos(3x)}{2 + \sin(3x)}

By the chain rule,

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

, where

u = 2 + \sin(3x)

and

\frac{du}{dx} = 3\cos(3x)

Differentiate the exponential component

e^{4x}

\frac{d}{dx}[e^{4x}] = 4e^{4x}

The standard rule for exponential differentiation states that

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

Combine the results to state the derivative function

\frac{dy}{dx}

\frac{dy}{dx} = \frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}

The derivative of a sum of functions is the sum of their individual derivatives.

Evaluate

\frac{dy}{dx}

x = 0

\frac{dy}{dx}\Big|_{x=0} = \frac{3\cos(0)}{2 + \sin(0)} + 4e^0 = \frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5

Substituting

x = 0

uses the values

\cos(0) = 1

\sin(0) = 0

, and

e^0 = 1

Differentiation of Logarithmic, Trigonometric, and Exponential Functions using the Chain Rule

Question