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

If $y = x^3 \sin(2x)$ , what is $\frac{dy}{dx}$ ?

The derivative

\frac{dy}{dx}

3x^2 \sin(2x) + 2x^3 \cos(2x)

Applying the product rule

\frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx}

with

u = x^3

and

v = \sin(2x)

yields

x^3(2\cos(2x)) + \sin(2x)(3x^2)

, which simplifies to

3x^2 \sin(2x) + 2x^3 \cos(2x)

Identify the two component functions for the Product Rule

Let

u = x^3

and

v = \sin(2x)

, so

y = uv

The function is a product of an algebraic term and a trigonometric term.

Find the derivative of each component function

\frac{du}{dx} = 3x^2 and \frac{dv}{dx} = 2\cos(2x) using the chain rule for \sin(2x).

The derivative of \sin(2x) with respect to 2x is \cos(2x), multiplied by the derivative of 2x which is 2.

Apply the Product Rule formula \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

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

Combining the differentiated terms according to the product rule gives the final derivative.

Product Rule combined with Chain Rule for trigonometric functions

Estimated Time:1m 30s

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