If y = e^{3x}\cos(2x) + \ln(x + 1), calculate the value of \frac{dy}{dx} at x = 0. — JAMB UTME

If $y = e^{3x}\cos(2x) + \ln(x + 1)$ , calculate the value of $\frac{dy}{dx}$ at $x = 0$ .

Answer: 4

The derivative evaluated at

x = 0

is 4.

Differentiating

y = e^{3x}\cos(2x) + \ln(x + 1)

with respect to

x

yields

\frac{dy}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x) + \frac{1}{x+1}

. Evaluating this derivative at

x = 0

gives

3(1)(1) - 2(1)(0) + 1 = 4

Differentiate the product

u(x) = e^{3x}\cos(2x)

using the product rule and chain rule.

\frac{du}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x)

By the product rule

\frac{d}{dx}(uv) = u'v + uv'

, where

\frac{d}{dx}(e^{3x}) = 3e^{3x}

and

\frac{d}{dx}(\cos(2x)) = -2\sin(2x)

Differentiate the logarithmic term

v(x) = \ln(x + 1)

\frac{dv}{dx} = \frac{1}{x + 1}

The derivative of

\ln(g(x))

\frac{g'(x)}{g(x)}

Combine the terms to write the complete derivative

\frac{dy}{dx}

\frac{dy}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x) + \frac{1}{x + 1}

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} = 3e^0\cos(0) - 2e^0\sin(0) + \frac{1}{0 + 1} = 3(1)(1) - 2(1)(0) + 1 = 4

Substitute

x = 0

using

e^0 = 1

\cos(0) = 1

, and

\sin(0) = 0

Differentiation of Trigonometric, Exponential, and Logarithmic Functions

Estimated Time:1m 30s

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