Problem 17 - Calculus Early Transcendentals

Question

a. Use the Product Rule to find the derivative of the given function. Simplify your result. b. Find the derivative by expanding the product first. Verify that your answer agrees with part \((a)\) $$g(y)=\left(3 y^{4}-y^{2}\right)\left(y^{2}-4\right)$$

Step-by-Step Solution

Answer

Question: Verify that the derivative \(g'(y)\) obtained by applying the Product Rule and by expanding the product first are the same for the given function \(g(y) = (3y^{4} - y^{2})(y^{2} - 4)\). Answer: The derivative \(g'(y)\) for both methods is found to be \(18y^{5} - 52y^{3} + 8y\).

Step 1: Apply the Product Rule

Let's find the derivative of \(g(y)\) using the Product Rule. The Product Rule states that for two functions \(u(y)\) and \(v(y)\), the derivative \((u(y)v(y))' = u'(y)v(y) + u(y)v'(y)\). In our case, \(u(y) = 3y^{4} - y^{2}\) and \(v(y) = y^{2} - 4\). First, we compute the derivatives of \(u(y)\) and \(v(y)\): $$u'(y)=(3y^{4} - y^{2})' = 12y^{3} - 2y$$ $$v'(y)=(y^{2} - 4)' = 2y$$ Now, apply the Product Rule: $$g'(y) = u'(y)v(y) + u(y)v'(y) = (12y^{3} - 2y)(y^{2} - 4) + (3y^{4} - y^{2})(2y)$$

Step 2: Expand the product first

To find the derivative by expanding the product first, we will multiply \(u(y)\) and \(v(y)\): $$g(y) = (3y^{4} - y^{2})(y^{2} - 4) = 3y^{6} - 12y^{4} - y^{4} + 4y^{2} = 3y^{6} - 13y^{4} + 4y^{2}$$ Now, compute the derivative of the expanded function \(g(y)\): $$g'(y) = (3y^{6} - 13y^{4} + 4y^{2})' = 18y^{5} - 52y^{3} + 8y$$

Step 3: Simplify and compare the results

We now simplify the expression we obtained in Step 1: $$g'(y) = (12y^{3} - 2y)(y^{2} - 4) + (3y^{4} - y^{2})(2y)$$ $$g'(y) = 12y^{5} - 48y^{3} - 2y^{3} + 8y + 6y^{5} - 2y^{3}$$ $$g'(y) = 18y^{5} - 52y^{3} + 8y$$ Comparing the results from Step 1 and Step 2, we can see that they are the same: $$g'(y) = 18y^{5} - 52y^{3} + 8y$$ Thus, both methods give us the same derivative, and the exercise is complete.

Key Concepts

Derivative Calculation

The process of derivative calculation is essentially finding the rate at which a function's output changes with respect to its input. This is a fundamental concept in calculus used to understand how different functions behave and is crucial for solving various practical problems in fields such as physics, engineering, and economics.

When dealing with product functions, where one function is multiplied by another, the Product Rule is indispensable. It states that the derivative of a product of two functions is the derivative of the first function multiplied by the second, plus the first function multiplied by the derivative of the second function. Mathematically, this can be noted as \[ (u(y)v(y))' = u'(y)v(y) + u(y)v'(y) \].

In the given exercise, we applied this rule to the function \(g(y) = (3y^4 - y^2)(y^2 - 4)\). The derivatives of \(u(y) = 3y^4 - y^2\) and \(v(y) = y^2 - 4\) were first calculated as \(12y^3 - 2y\) and \(2y\), respectively. Employing the Product Rule gave us a new function representing the rate of change of \(g(y)\) with respect to \(y\).

Calculus Simplification

In calculus, simplification involves streamlining expressions to make them easier to read or to prepare them for further operations. It can include distributing products, combining like terms, and factoring, among other techniques. Simplification makes it easier to compare results and to apply additional calculus tools on functions.

In our exercise, after using the Product Rule to find the derivative, we were left with a somewhat complex expression. This included distributing the products and combining like terms, resulting in the simplified final derivative \(g'(y) = 18y^5 - 52y^3 + 8y\). This clean expression readily shows the relationship between \(y\) and the derivative of \(g\), making it much easier to work with for further analysis or applications.

Function Expansion

Function expansion is a technique that involves manipulating a function to express it as a sum of terms, rather than a product of terms or a composition of functions. This expansion can often make certain types of calculations, especially derivatives, more straightforward.

In our exercise's alternate method, expanding \(g(y)\) before taking the derivative—effectively abusing the distributive property of multiplication—led to \(g(y) = 3y^6 - 13y^4 + 4y^2\). This expansion made the derivative more apparent because we could directly apply the power rule to each term separately, which states that the derivative of \( y^n \) is \( ny^{n-1} \).

Ultimately, function expansion confirmed the result from the Product Rule, showing that \(g'(y) = 18y^5 - 52y^3 + 8y\) regardless of the method. Both function expansion and calculus simplification are effective strategies for tackling complex problems. Choosing one over the other may depend on the specific situation and the preferences of the person doing the calculation.