Question

Differentiate: \( y = x^2\sin x \)

• A. \( (2x\sin x) \)
• B. \( (x^2\cos x) \)
• C. \( (2x\sin x + x^2\cos x) \)
• D. \( (2x\cos x) \)

Hint:

Always identify the structure of the function (e.g., product, quotient, chain rule) before differentiating. The product rule is crucial for functions formed by multiplying two variable expressions.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to differentiate the given function \( y = x^2\sin x \) with respect to \( x \).

Step 2: Key Formula or Approach:
The function is a product of two functions of \( x \): \( u(x) = x^2 \) and \( v(x) = \sin x \). Therefore, we need to use the product rule of differentiation.
Product Rule: If \( y = u(x)v(x) \), then \( \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \).
Also, remember basic differentiation rules:
- \( \frac{d}{dx}(x^n) = nx^{n-1} \)
- \( \frac{d}{dx}(\sin x) = \cos x \)

Step 3: Detailed Explanation:
Given function: \( y = x^2\sin x \)
Let \( u(x) = x^2 \) and \( v(x) = \sin x \).
Calculate the derivatives of \( u(x) \) and \( v(x) \):
- \( u'(x) = \frac{d}{dx}(x^2) = 2x \)
- \( v'(x) = \frac{d}{dx}(\sin x) = \cos x \)
Now, apply the product rule:
\[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \]
\[ \frac{dy}{dx} = (2x)(\sin x) + (x^2)(\cos x) \]
\[ \frac{dy}{dx} = 2x\sin x + x^2\cos x \]
This matches option (C).

Step 4: Final Answer:
The derivative is \( 2x\sin x + x^2\cos x \).