Question

The derivative of x$^2$ cos x is _____________

• A. $2x \sin x - x^2 \sin x$
• B. $2x \cos x - x^2 \sin x$
• C. $2x \sin x - x^2 \cos x$
• D. $\cos x - x^2 \sin x \cos x$

Hint:

Always remember the negative sign when differentiating cosine:
- \(\frac{d}{dx}(\cos x) = -\sin x\)
Using the product rule, this negative sign is placed in front of the second term:
\[ \text{Derivative} = (\text{Derivative of first}) \cdot (\text{second}) + (\text{first}) \cdot (\text{Derivative of second}) \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the derivative of a product of two functions, we must apply the Product Rule of differentiation.
Key Formula or Approach:
The Product Rule states that if \(y = u(x) \cdot v(x)\), then its derivative is:
\[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \] The standard derivatives used here are:
\[ \frac{d}{dx}(x^2) = 2x \] \[ \frac{d}{dx}(\cos x) = -\sin x \]

Step 2: Detailed Explanation:
Let the function be defined as:
\[ y = x^2 \cos x \] Identify the two components:
- \(u(x) = x^2\)
- \(v(x) = \cos x\)
Now, find their individual derivatives:
- \(u'(x) = 2x\)
- \(v'(x) = -\sin x\)
Substitute these terms into the product rule formula:
\[ \frac{dy}{dx} = (2x)(\cos x) + (x^2)(-\sin x) \] \[ \frac{dy}{dx} = 2x \cos x - x^2 \sin x \] This matches Option (B).

Step 3: Final Answer:
The derivative of \(x^2 \cos x\) is \(2x \cos x - x^2 \sin x\).