Answer:
![\displaystyle \large{f\prime(x) = 2x \sin (x) + x^2 \cos (x)}](https://img.qammunity.org/2019/formulas/mathematics/high-school/b3my91u0240nrm8qt8olue63jjxfhhiurq.png)
Explanation:
We are given a function:
![\displaystyle \large{f(x) = x^2 \sin (x)}](https://img.qammunity.org/2019/formulas/mathematics/high-school/po4bejh8tqs7mdie761a87u02mhfjejhuc.png)
Notice that there are two functions multiplying each other. Recall the product rules.
![\displaystyle \large{y=h(x)g(x) \to y\prime = h\prime (x) g(x) + h(x)g\prime (x)}](https://img.qammunity.org/2019/formulas/mathematics/high-school/d5d5z6rcwgl7550s8az9fvdhiai3xwrala.png)
You can let x^2 = h(x), sin(x) = g(x) or sin(x) = h(x), x^2 = g(x) as your desire but I’ll let h(x) = x^2 and g(x) = sin(x).
Therefore, from a function:
![\displaystyle \large{f(x) = x^2 \sin (x) \to f\prime (x) = (x^2)\prime \sin(x) + x^2 (\sin (x))\prime}](https://img.qammunity.org/2019/formulas/mathematics/high-school/ovknormbkgw6ppb9rv838z7qxoqgp4j8fl.png)
Recall the power rules and differentiation of sine.
![\displaystyle \large{y = ax^n \to y\prime = nax^(n-1) \ \ \ \tt{for \ \ polynomial \ \ function}}\\ \displaystyle \large{y = \sin (x) \to y\prime = \cos (x) }](https://img.qammunity.org/2019/formulas/mathematics/high-school/et63bvmzq5qejoyo4xsv2oxicidlczw7q9.png)
Therefore, from differentiating function,
![\displaystyle \large{f\prime(x) = 2x \sin (x) + x^2 \cos (x)}](https://img.qammunity.org/2019/formulas/mathematics/high-school/b3my91u0240nrm8qt8olue63jjxfhhiurq.png)
And we are done!