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

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

Answer: 10

Differentiating

y = 2\sin(3x) + e^{4x}

yields

\frac{dy}{dx} = 6\cos(3x) + 4e^{4x}

. Evaluating this expression at

x = 0

gives

6\cos(0) + 4e^{0} = 6(1) + 4(1) = 10

Differentiate each term of

y = 2\sin(3x) + e^{4x}

with respect to

x

\frac{dy}{dx} = 6\cos(3x) + 4e^{4x}

Applying the chain rule gives

\frac{d}{dx}[2\sin(3x)] = 2 \times 3\cos(3x) = 6\cos(3x)

and

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

Evaluate the derivative at

x = 0

6\cos(0) + 4e^{0} = 6(1) + 4(1) = 10

Substituting

x = 0

gives

\cos(0) = 1

and

e^{0} = 1

Differentiation of trigonometric and exponential functions

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