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

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

Answer: 5

Evaluating the second derivative of

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

x = 0

yields

5

. This is calculated by applying the product, chain, and quotient rules to get

\frac{d^2 y}{dx^2}

, and substituting

x = 0

, where

e^0 = 1

\sin 0 = 0

\cos 0 = 1

, and

\ln 1 = 0

Differentiate

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

with respect to

x

using the product rule.

\frac{dy}{dx} = 2e^{2x} \ln(e^x + \sin x) + e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

The function is composed of

u(x) = e^{2x}

and

v(x) = \ln(e^x + \sin x)

. By chain rule,

v'(x) = \frac{e^x + \cos x}{e^x + \sin x}

Differentiate the first term

T_1(x) = 2e^{2x} \ln(e^x + \sin x)

to get

T_1'(x)

T_1'(x) = 4e^{2x} \ln(e^x + \sin x) + 2e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

Applying the product rule to

2e^{2x}

and

\ln(e^x + \sin x)

Differentiate the second term

T_2(x) = e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

using the product and quotient rules.

T_2'(x) = 2e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right) + e^{2x} \left(\frac{(e^x - \sin x)(e^x + \sin x) - (e^x + \cos x)^2}{(e^x + \sin x)^2}\right)

The derivative of

\frac{e^x + \cos x}{e^x + \sin x}

requires the quotient rule.

Evaluate

T_1'(0)

and

T_2'(0)

x = 0

T_1'(0) = 4(1)(0) + 2(1)(2) = 4

, and

T_2'(0) = 2(1)(2) + 1 \left(\frac{(1)(1) - 2^2}{1^2}\right) = 4 - 3 = 1

x = 0

e^0 = 1

\sin 0 = 0

\cos 0 = 1

, and

\ln(1) = 0

Sum the evaluated derivative components to find

\frac{d^2 y}{dx^2}\Big|_{x=0}

\frac{d^2 y}{dx^2}\Big|_{x=0} = T_1'(0) + T_2'(0) = 4 + 1 = 5

Combining the evaluated terms gives the final numerical value.

Higher-Order Derivatives of Combined Transcendental Functions

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