Question

question determine $h(x)$ when $h(x)=0.04e^{x}x^{2}$. select the correct answer below: $h(x)=0.04(e^{x})(2x)$ $h(x)=0.04((xe^{x - 1})(x^{2})+(e^{x})(2x))$ $h(x)=0.04((e^{x})(x^{2})+(e^{x})(2x))$ $h(x)=0.04(e^{x})(x^{2})$

Explanation:

Step1: Apply product - rule

The product - rule states that if $h(x)=u(x)v(x)$, then $h'(x)=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=0.04e^{x}$ and $v(x)=x^{2}$. Then $u'(x)=0.04e^{x}$ and $v'(x)=2x$.

Step2: Calculate $h'(x)$

$h'(x)=0.04e^{x}(2x)+(0.04e^{x})(x^{2})$.

Answer:

$h'(x)=0.04((e^{x})(2x)+(e^{x})(x^{2}))$