Problem 1 - Calculus and its applications

Question

Differentiate two ways: first, by using the Product Rule; then, by multiplying the expressions before differentiating. Compare your results as a check. Use a graphing calculator to check your results. \(y=x^{9} \cdot x^{4}\)

Step-by-Step Solution

Answer

Both methods yield the derivative \(13x^{12}\).

Step 1: Differentiate using the Product Rule

The Product Rule states that if you have a product of two functions, such as \( u(x) \) and \( v(x) \), then the derivative is given by \((u'v + uv')\). Here, let both \( u(x) \) and \( v(x) \) be \(x^9\) and \(x^4\) respectively. \(y = x^9 \cdot x^4\) can split as \(u(x) = x^9\) and \(v(x) = x^4\). Now find the derivatives of both functions. For \( u(x) = x^9, \ u'(x) = 9x^8 \).For \( v(x) = x^4, \ v'(x) = 4x^3 \).Substitute into the Product Rule formula:\(y' = u'v + uv' = (9x^8)(x^4) + (x^9)(4x^3)\).Calculate: \(y' = 9x^{12} + 4x^{12}\).Combine like terms: \(y' = 13x^{12}\).

Step 2: Simplify Before Differentiating

Before using derivatives, simplify the function by combining the exponents:\(y = x^9 \cdot x^4 = x^{9+4} = x^{13}\).Now differentiate with respect to \(x\): \(y' = 13x^{12}\).

Step 3: Compare the Results

Both methods give the same derivative:From the Product Rule: \(y' = 13x^{12}\).From simplifying first: \(y' = 13x^{12}\).Both methods confirm the derivative is correct.

Step 4: Verify with a Graphing Calculator

Enter the original formula for \(y = x^9 \cdot x^4\) in a graphing calculator and graph its derivative. Check that the calculator's result is consistent with our analytical solution, showing the derivative as \(13x^{12}\).

Key Concepts

Product Rule

When faced with finding the derivative of a product of two functions, such as in the expression \( y = x^9 \cdot x^4 \), the Product Rule is a handy tool. It gives us a method to differentiate products by separating them into simpler components.
In general, the Product Rule states: if you have a function \( y = u(x) \cdot v(x) \), the derivative \( y' \) will be the sum of the derivative of \( u(x) \) times \( v(x) \) plus \( u(x) \) times the derivative of \( v(x) \).

This is written as:

• \( y' = u'(x)v(x) + u(x)v'(x) \)

Apply this to our expression by treating \( x^9 \) as \( u(x) \) and \( x^4 \) as \( v(x) \). Differentiate each:

• \( u'(x) = 9x^8 \)
• \( v'(x) = 4x^{3} \)

Plug these derivatives back into the formula:

• \( y' = 9x^8 \cdot x^4 + x^9 \cdot 4x^3 \)

Calculating this gives \( y' = 13x^{12} \), confirming the derivative using the Product Rule.

Differentiation

Differentiation is the process of finding the derivative of a function. It tells us how the function changes, or its rate of change, with respect to its variable.
In the context of our exercise, differentiation helps us understand how \( y = x^9 \cdot x^4 \) changes as \( x \) changes.

When differentiating, it's key to recognize patterns and methods that simplify the process. For products of simple power functions like \( x^9 \cdot x^4 \), we can apply differentiation rules such as:

• Product Rule
• Power Rule

Here, the derivation involved using both the Product Rule and Power Rule effectively. Differentiation boils down to applying these rules correctly and ensuring that each step logically follows the last, yielding accurate results in understanding how the function behaves.

Power Rule

The Power Rule is a straightforward rule that simplifies taking derivatives of functions in the form \( x^n \). It states that if \( y = x^n \), then the derivative \( y' = nx^{n-1} \).

In our example, simplifying \( y = x^9 \cdot x^4 \) first results in \( y = x^{13} \). Using the Power Rule, differentiating \( x^{13} \) becomes clear:

• \( y' = 13x^{12} \)

The Power Rule allows us to quickly find the derivative by multiplying the coefficient by the power of \( x \), then reducing that power by one.
This provides an elegant solution, especially when combined with other rules, like the Product Rule in more complex expressions.