Question
The gradient of $y = 3x^2 \sin(2x)$ at (0.2, 1) is __________ (rounded off to three decimal places).
The function given is \( y = 3x^2 \sin(2x) \). To find the gradient at the point (0.2, 1), we need to compute the derivative of \( y \) with respect to \( x \) and evaluate it at \( x = 0.2 \).
First, use the product rule for differentiation, which states that if \( u(x) \) and \( v(x) \) are two functions, then the derivative of their product \( y = u(x)v(x) \) is given by:
$$ y' = u'v + uv' $$
In our case, let \( u(x) = 3x^2 \) and \( v(x) = \sin(2x) \).
The derivative of \( u(x) \) is:
$$ u' = \frac{d}{dx}(3x^2) = 6x $$
The derivative of \( v(x) \) is:
$$ v' = \frac{d}{dx}(\sin(2x)) = 2\cos(2x) $$
Using the product rule, the derivative of \( y \) is:
$$ y' = 6x\sin(2x) + 3x^2(2\cos(2x)) $$
Simplifying, we get:
$$ y' = 6x\sin(2x) + 6x^2\cos(2x) $$
Now evaluate \( y' \) at \( x = 0.2 \):
$$ y'(0.2) = 6(0.2)\sin(2 \times 0.2) + 6(0.2)^2\cos(2 \times 0.2) $$
Calculating further:
$$ = 1.2\sin(0.4) + 0.24\cos(0.4) $$
Using approximate values: \(\sin(0.4) \approx 0.389\) and \(\cos(0.4) \approx 0.921\), then:
$$ y'(0.2) = 1.2 \times 0.389 + 0.24 \times 0.921 $$
$$ = 0.4668 + 0.22104 $$
$$ = 0.68784 $$
Rounded to three decimal places, the gradient at (0.2, 1) is 0.688.
Since 0.688 falls within the expected range (0.67, 0.7), the computation is confirmed correct. Therefore, the gradient of \( y = 3x^2 \sin(2x) \) at (0.2, 1) is \( 0.688 \).
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