If y = e^{2x} \cos 3x, what is \frac{dy}{dx}? — JAMB UTME

If $y = e^{2x} \cos 3x$ , what is $\frac{dy}{dx}$ ?

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

Applying the product rule to

u = e^{2x}

and

v = \cos 3x

yields

\frac{du}{dx} = 2e^{2x}

and

\frac{dv}{dx} = -3\sin 3x

. Substituting these into

u\frac{dv}{dx} + v\frac{du}{dx}

gives

e^{2x}(-3\sin 3x) + \cos 3x(2e^{2x}) = e^{2x}(2\cos 3x - 3\sin 3x)

Identify the components for the product rule

Let

u = e^{2x}

and

v = \cos 3x

The function

y = e^{2x} \cos 3x

is a product of two differentiable functions.

Differentiate

u = e^{2x}

with respect to

x

\frac{du}{dx} = 2e^{2x}

By the chain rule,

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

Differentiate

v = \cos 3x

with respect to

x

\frac{dv}{dx} = -3\sin 3x

By the chain rule,

\frac{d}{dx}(\cos kx) = -k \sin kx

Apply the product rule formula

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

and factor out

e^{2x}

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

Combining the products and factoring out the common exponential factor simplifies the expression.

Product Rule and Chain Rule for Exponential and Trigonometric Functions

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