Question:
The gradient of \( y = 3x^2 \sin(2x) \) at \( (0.2, 1) \) is:
The gradient of \( y = 3x^2 \sin(2x) \) at \( (0.2, 1) \) is:
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To find the gradient of a function at a specific point, differentiate the function and substitute the given \( x \)-coordinate into the derivative.
Updated On: Apr 25, 2025
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Solution and Explanation
Step 1: Differentiate the function \( y = 3x^2 \sin(2x) \) using the product rule.
The product rule states that: \[ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \] Let \( u(x) = 3x^2 \) and \( v(x) = \sin(2x) \). Thus, \[ u'(x) = 6x, \quad v'(x) = 2\cos(2x) \] Now, applying the product rule: \[ \frac{dy}{dx} = 6x \sin(2x) + 3x^2 \cdot 2\cos(2x) = 6x \sin(2x) + 6x^2 \cos(2x) \] Step 2: Evaluate the derivative at \( x = 0.2 \).
Substitute \( x = 0.2 \) into the expression for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = 6(0.2) \sin(2 \times 0.2) + 6(0.2)^2 \cos(2 \times 0.2) \] \[ = 1.2 \sin(0.4) + 6(0.04) \cos(0.4) \] \[ = 1.2 \times 0.3894 + 0.24 \times 0.9218 \] \[ = 0.4673 + 0.2212 \] \[ = 0.6885 \] Final Answer: The gradient at \( (0.2, 1) \) is approximately \( 0.689 \).
The product rule states that: \[ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \] Let \( u(x) = 3x^2 \) and \( v(x) = \sin(2x) \). Thus, \[ u'(x) = 6x, \quad v'(x) = 2\cos(2x) \] Now, applying the product rule: \[ \frac{dy}{dx} = 6x \sin(2x) + 3x^2 \cdot 2\cos(2x) = 6x \sin(2x) + 6x^2 \cos(2x) \] Step 2: Evaluate the derivative at \( x = 0.2 \).
Substitute \( x = 0.2 \) into the expression for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = 6(0.2) \sin(2 \times 0.2) + 6(0.2)^2 \cos(2 \times 0.2) \] \[ = 1.2 \sin(0.4) + 6(0.04) \cos(0.4) \] \[ = 1.2 \times 0.3894 + 0.24 \times 0.9218 \] \[ = 0.4673 + 0.2212 \] \[ = 0.6885 \] Final Answer: The gradient at \( (0.2, 1) \) is approximately \( 0.689 \).
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