[-/1 Points] DETAILS MY NOTES HARMATHAP12 9.4.050. The monthly output of a certain product is \[ Q(x)=2300 x^{5 / 2} \] mhere \( x \) is the capital investhent in milions of dollars. Find dQidx, which can be used to estimate the effect on the output if an additional capital investment of \( \$ 1 \) milion is made. \[ d Q / d x= \] \( \square \) Need Help? Paad 12 Farch 1 SUBMIT ANSWER

To find \(\frac{dQ}{dx}\), we need to differentiate the function \(Q(x) = 2300x^{5/2}\) with respect to \(x\). Applying the power rule, we have: \[ \frac{dQ}{dx} = 2300 \cdot \frac{5}{2} x^{3/2} = 5750 x^{3/2} \] This derivative \(\frac{dQ}{dx}\) tells us the rate of change of output with respect to capital investment. So whenever you invest an additional million dollars, multiply \(5750\) by \(x^{3/2}\) (where \(x\) is the current investment) to estimate the increase in output! Let’s look at some real-world applications! In industries where capital investment is essential, knowing how output changes with investment is invaluable. For instance, in manufacturing, a company can predict how much additional revenue it could generate per million invested, helping to make better financial decisions. Using derivatives helps businesses optimize their capital allocation and forecast growth potential accurately!