If y = e^{-2x} \sin(3x), find \frac{dy}{dx}. — JAMB UTME

If $y = e^{-2x} \sin(3x)$ , find $\frac{dy}{dx}$ .

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

Applying the product rule

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

u = e^{-2x}

and

v = \sin(3x)

yields

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

and

\frac{dv}{dx} = 3\cos(3x)

. Substituting these terms gives

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

Identify the component functions for the product rule

Let

u = e^{-2x}

and

v = \sin(3x)

The function

y

is a product of an exponential function and a trigonometric function.

Differentiate each component using the chain rule

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

and

\frac{dv}{dx} = 3\cos(3x)

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

and

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

where

k

is a constant.

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

Combining terms correctly gives the exact derivative.

Differentiation of Product of Transcendental Functions (Exponential and Trigonometric)

Soru