(1 point) Differentiate g(x)=11sqrt (x)cdot e^x Answer: g'(x)=

To differentiate the function \( g(x) = 11\sqrt{x} \cdot e^x \), we will use the product rule. The product rule states that if you have a function \( h(x) = f(x) \cdot g(x) \), then its derivative is given by:<br /><br />\[ h'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) \]<br /><br />Here, let \( f(x) = 11\sqrt{x} \) and \( g(x) = e^x \).<br /><br />First, we need to find the derivatives of \( f(x) \) and \( g(x) \):<br /><br />1. \( f(x) = 11\sqrt{x} = 11x^{1/2} \)<br /> \[ f'(x) = 11 \cdot \frac{1}{2} x^{-1/2} = \frac{11}{2} x^{-1/2} = \frac{11}{2\sqrt{x}} \]<br /><br />2. \( g(x) = e^x \)<br /> \[ g'(x) = e^x \]<br /><br />Now, apply the product rule:<br /><br />\[ g'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) \]<br />\[ g'(x) = \left( \frac{11}{2\sqrt{x}} \right) \cdot e^x + 11\sqrt{x} \cdot e^x \]<br /><br />Combine the terms:<br /><br />\[ g'(x) = \frac{11e^x}{2\sqrt{x}} + 11\sqrt{x}e^x \]<br /><br />To combine these terms, express them with a common denominator:<br /><br />\[ g'(x) = \frac{11e^x}{2\sqrt{x}} + \frac{22x e^x}{2\sqrt{x}} \]<br />\[ g'(x) = \frac{11e^x + 22xe^x}{2\sqrt{x}} \]<br />\[ g'(x) = \frac{11e^x(1 + 2x)}{2\sqrt{x}} \]<br /><br />Thus, the derivative of \( g(x) = 11\sqrt{x} \cdot e^x \) is:<br /><br />\[ g'(x) = \frac{11e^x(1 + 2x)}{2\sqrt{x}} \]