Since the units on \(AC\) are dollars per item, and the units on \(x\) are thousands of items, the units on \(AC'\) are dollars per item per thousands of items. ![]() The total volume of water in the tank after \(t\) minutes is \(50 + 5t\), so the concentration after \(t\) minutes is \ The number of kg of chemical stays constant at 8 kg, but the quantity of water in the tank is increasing by 5 L/min. The concentration would be measured as kg of chemical per liter of water, kg/L. If a tap is opened and water is added to the tank at a rate of 5 liters per minute, at what rate is the concentration of chemical in the tank changing after 4 minutes?įirst we need to set up a model for the concentration of chemical. Suppose a large tank contains 8 kg of a chemical dissolved in 50 liters of water. We can use these rules, together with the basic rules, to find derivatives of many complicated looking functions. They also let us deal with products where the factors are not polynomials. The rules for finding derivatives of products and quotients are a little complicated, but they save us the much more complicated algebra we might face if we were to try to multiply things out. What if we try differentiating the factors and multiplying them? We’d get \( h'(x)=\left(12x^2\right)(1)=12x^2 \), which is radically different from the correct answer. We already worked out the derivative, it is \( h'(x)=16x^3-11+36x^2 \). To see that, consider finding derivative of \( h(x)=\left(4x^3-11\right)(x+3) \). It would be great if we can just take the derivatives of the factors and multiply them, but unfortunately that won’t give the right answer. We’ll need a rule for finding the derivative of a product so we don’t have to multiply everything out. We could 'simply' multiply it out to find its derivative as before – who wants to volunteer? Nobody? This function is not a simple sum or difference of polynomials. Now suppose we wanted to find the derivative of \ We can simply multiply it out to find its derivative:
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