By the definition of derivative, \[(fg)'(a) = \lim_{h \rightarrow 0} \frac{f(a+h)g(a+h) - f(a)g(a)}{h}\] Add and subtract \(f(a)g(a+h),\) then factor. \begin{align} \lim_{h \rightarrow 0} \frac{f(a+h)g(a+h) - f(a)g(a)}{h} & = \lim_{h \rightarrow 0} \frac{f(a+h)g(a+h) - f(a)g(a+h) + f(a)g(a+h) - f(a)g(a)}{h} \\ & = \lim_{h \rightarrow 0} \frac{(f(a+h) - f(a))g(a+h) + f(a)(g(a+h) - g(a))}{h} \\ & = \lim_{h \rightarrow 0} \frac{(f(a+h) - f(a))g(a+h)}{h} + \lim_{h \rightarrow 0} \frac{f(a)(g(a+h) - g(a))}{h} \end{align} Now use \(\lim_{h \rightarrow 0} g(a+h) = g(a)\) and the definition of the derivatives of \(f\) and \(g\) to get \[\lim_{h \rightarrow 0} \frac{(f(a+h) - f(a))g(a+h)}{h} + \lim_{h \rightarrow 0} \frac{f(a)(g(a+h) - g(a))}{h} = f'(a)g(a)+f(a)g'(a)\]

By the power rule, the derivative of \(x^2\) is \(2x.\) The derivative of \(\text{sin}(x)\) is \(\text{cos}(x).\) So, by the product rule, \[\frac{d}{dx}x^2\text{sin}(x) = 2x\text{sin}(x) + x^2\text{cos}(x)\]

By the power rule, the derivative of \(x^3\) is \(3x^2.\) The derivative of \(e^x\) is \(e^x.\) So, by the product rule, \[\frac{d}{dx}x^3e^x = 3x^2e^x + x^3e^x\]

The derivative of \(e^x\) is \(e^x\) and the derivative of \(\text{sin}(x)\) is \(\text{cos}(x).\) So, by the product rule, \[\frac{d}{dx}e^x\text{sin}(x) = e^x\text{sin}(x) + e^x\text{cos}(x)\]