Question

1. - / 5.26 points

find the first and second derivatives of the function.
y = sin(x²)
y =
y =
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Explanation:

Step1: Compute first derivative using chain rule

Let $u = x^2$, so $y = \sin(u)$. Then $y' = \cos(u) \cdot u' = \cos(x^2) \cdot 2x = 2x\cos(x^2)$

Step2: Compute second derivative using product rule

$y'' = \frac{d}{dx}[2x\cos(x^2)] = 2\cos(x^2) + 2x \cdot [-\sin(x^2) \cdot 2x] = 2\cos(x^2) - 4x^2\sin(x^2)$

Answer:

$y' = 2x\cos(x^2)$
$y'' = 2\cos(x^2) - 4x^2\sin(x^2)$