Question: The derivative of \( f(x) = x^4 \sin(x) \) is \( \frac{d}{dx} f(x) = -x^4 \cos(x) + 4x^3…

The derivative of \( f(x) = x^4 \sin(x) \) is \( \frac{d}{dx} f(x) = -x^4 \cos(x) + 4x^3 \sin(x) \).

True False

Give your reasons

Solution

The problem asks if the derivative of \( f(x) = x^4 \sin(x) \) is \( \frac{d}{dx} f(x) = -x^4 \cos(x) + 4x^3 \sin(x) \). To solve, we’ll use the product rule for differentiation. The product rule states that if you have two functions, \( u(x) \) and \( v(x) \), then the derivative of their product is given by: \[ \frac{d}{dx}[u(x)v(x)] = u’(x)v(x) + u(x)v’(x) \] Let’s apply the product rule to \( f(x) = x^4 \sin(x) \). Set \( u(x) = x^4 \) and \( v(x) = \sin(x) \). First, find the derivatives: \( u’(x) = \frac{d}{dx}[x^4] = 4x^3 \), \( v’(x) = \frac{d}{dx}[\sin(x)] = \cos(x) \). Now apply the product rule: \[ \frac{d}{dx}[x^4 \sin(x)] = 4x^3 \sin(x) + x^4 \cos(x) \] The given expression is: \(-x^4 \cos(x) + 4x^3 \sin(x)\). Comparing both expressions, we see they are not equal. Therefore, the statement is indeed False.