If y = 2e^{4x} + \ln(2x + 1), what is the value of \frac{dy}{dx} at x = 0? — JAMB UTME

If $y = 2e^{4x} + \ln(2x + 1)$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Answer: 10

Differentiating

y = 2e^{4x} + \ln(2x + 1)

with respect to

x

yields

\frac{dy}{dx} = 8e^{4x} + \frac{2}{2x + 1}

. Evaluating this derivative at

x = 0

gives

8(e^0) + \frac{2}{2(0) + 1} = 8(1) + 2 = 10

Differentiate each term of

y = 2e^{4x} + \ln(2x + 1)

with respect to

x

\frac{dy}{dx} = 8e^{4x} + \frac{2}{2x + 1}

By the chain rule,

\frac{d}{dx}(ae^{kx}) = ak e^{kx}

and

\frac{d}{dx}(\ln(u(x))) = \frac{u'(x)}{u(x)}

Evaluate the derivative at

x = 0

\left.\frac{dy}{dx}\right|_{x=0} = 8e^{0} + \frac{2}{2(0) + 1} = 8(1) + \frac{2}{1} = 10

Substitute

x = 0

into the derived expression and simplify using

e^0 = 1

Differentiation of exponential and logarithmic functions using the chain rule

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