differentiate. r(t) = (t + e^t)(6 - √t) r(t) =

Explanation:

Step1: Apply product - rule

The product - rule states that if $R(t)=u(t)v(t)$, then $R^{\prime}(t)=u^{\prime}(t)v(t)+u(t)v^{\prime}(t)$. Let $u(t)=t + e^{t}$ and $v(t)=6-\sqrt{t}=6 - t^{\frac{1}{2}}$.

Step2: Differentiate $u(t)$

Differentiate $u(t)=t + e^{t}$ with respect to $t$. Using the sum - rule and basic derivative formulas, $u^{\prime}(t)=\frac{d}{dt}(t)+\frac{d}{dt}(e^{t})=1 + e^{t}$.

Step3: Differentiate $v(t)$

Differentiate $v(t)=6 - t^{\frac{1}{2}}$ with respect to $t$. Using the difference - rule and power - rule, $v^{\prime}(t)=\frac{d}{dt}(6)-\frac{d}{dt}(t^{\frac{1}{2}})=0-\frac{1}{2}t^{-\frac{1}{2}}=-\frac{1}{2\sqrt{t}}$.

Step4: Substitute into product - rule

$R^{\prime}(t)=(1 + e^{t})(6-\sqrt{t})+(t + e^{t})(-\frac{1}{2\sqrt{t}})$ $=(1 + e^{t})(6-\sqrt{t})-\frac{t + e^{t}}{2\sqrt{t}}$ $=6-\sqrt{t}+6e^{t}-e^{t}\sqrt{t}-\frac{\sqrt{t}}{2}-\frac{e^{t}}{2\sqrt{t}}$ $=6 + 6e^{t}-\frac{3}{2}\sqrt{t}-e^{t}\sqrt{t}-\frac{e^{t}}{2\sqrt{t}}$

Answer:

$6 + 6e^{t}-\frac{3}{2}\sqrt{t}-e^{t}\sqrt{t}-\frac{e^{t}}{2\sqrt{t}}$