compute the first two derivatives of $f(x)=xsin(x^{2})$.\na. $f(x)=$\nb. $f(x)=$

Explanation:

Step1: Use product - rule for first derivative

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Here, $u = x$ and $v=\sin(x^{2})$. The derivative of $u=x$ is $u^\prime = 1$, and for $v = \sin(x^{2})$, using the chain - rule (if $y=\sin(u)$ and $u = x^{2}$, then $\frac{dy}{dx}=\cos(u)\cdot2x=2x\cos(x^{2})$). So, $f^\prime(x)=1\cdot\sin(x^{2})+x\cdot2x\cos(x^{2})=\sin(x^{2}) + 2x^{2}\cos(x^{2})$.

Step2: Use product - rule for second derivative

$f^\prime(x)=\sin(x^{2})+2x^{2}\cos(x^{2})$. Let $y_1=\sin(x^{2})$ and $y_2 = 2x^{2}\cos(x^{2})$. For $y_1=\sin(x^{2})$, using the chain - rule, $y_1^\prime=2x\cos(x^{2})$. For $y_2 = 2x^{2}\cos(x^{2})$, using the product - rule with $u = 2x^{2}$, $u^\prime=4x$ and $v=\cos(x^{2})$, $v^\prime=-2x\sin(x^{2})$. Then $y_2^\prime=4x\cos(x^{2})+2x^{2}\cdot(-2x\sin(x^{2}))=4x\cos(x^{2})-4x^{3}\sin(x^{2})$. So, $f^{\prime\prime}(x)=2x\cos(x^{2})+4x\cos(x^{2})-4x^{3}\sin(x^{2})=6x\cos(x^{2})-4x^{3}\sin(x^{2})$.

Answer:

a. $f^\prime(x)=\sin(x^{2}) + 2x^{2}\cos(x^{2})$ b. $f^{\prime\prime}(x)=6x\cos(x^{2})-4x^{3}\sin(x^{2})$