Problem 4 - 5 Steps to A 5: AP Calculus AB 2021

Question

Find the derivative of each of the following functions. $$y=\frac{x^{2}}{5 x^{6}-1}$$

Step-by-Step Solution

Answer

Answer: The derivative of the function $$y=\frac{x^2}{5x^6 - 1}$$ is $$y' = \frac{-20x^7 - 2x}{(5x^6 - 1)^2}$$.

Step 1: Identify u and v

In the given function, $$y=\frac{x^{2}}{5 x^{6}-1}$$ we can identify u and v as follows: $$ u = x^2 $$ $$ v = 5x^6 - 1 $$

Step 2: Calculate the derivatives u' and v'

Using power rule, calculate the derivatives of u and v: The derivative of $$u=x^2$$ with respect to x is: $$ u' = 2x $$ The derivative of $$v=5x^6 - 1$$ with respect to x is: $$ v' = 30x^5 $$

Step 3: Apply the Quotient Rule

Now apply the quotient rule to find the derivative of y with respect to x, using u, v, u', and v': $$ y' = \frac{u'v - uv'}{v^2} $$ Substituting the values for u, v, u', and v', we get: $$ y' = \frac{(2x)(5x^6 - 1) - (x^2)(30x^5)}{(5x^6 - 1)^2} $$

Step 4: Simplify the expression

Now let's simplify the expression: $$ y' = \frac{10x^7 - 2x - 30x^7}{(5x^6 - 1)^2} $$ Combine the terms in the numerator: $$ y' = \frac{-20x^7 - 2x}{(5x^6 - 1)^2} $$ The derivative of the given function, $$y=\frac{x^{2}}{5 x^{6}-1}$$ is: $$ y' = \frac{-20x^7 - 2x}{(5x^6 - 1)^2} $$

Key Concepts

Quotient Rule

When dealing with derivatives of a function presented as a quotient, such as in the form \( y = \frac{u}{v} \), the Quotient Rule is your go-to technique. Calculating the derivative of these types of functions requires more steps compared to simpler rules. The Quotient Rule states that for functions \( u(x) \) and \( v(x) \), the derivative \( y' \) is calculated as:

• \( y' = \frac{u'v - uv'}{v^2} \)

Where:

• \( u' \) is the derivative of the numerator.
• \( v' \) is the derivative of the denominator.
• \( v^2 \) represents the square of the denominator function.

This rule elegantly accounts for changes in both the numerator and denominator, keeping track of the interdependency between the two parts of the function.

Power Rule

The Power Rule is a straightforward yet powerful tool in calculus for finding derivatives of functions with powers of x. When you're trying to differentiate a function like \( u = x^n \), you apply the Power Rule: \( u' = nx^{n-1} \)This rule tells us to multiply the power \( n \) by \( x \) raised to the power of \( n-1 \). In our example exercise, we used the Power Rule to find that:

• The derivative of \( u = x^2 \) is \( u' = 2x \)
• The derivative of \( v = 5x^6 - 1 \) becomes \( v' = 30x^5 \)

Recognizing when and how to use the Power Rule makes solving more complex derivatives, such as those involving the Quotient Rule, more manageable.

Simplifying Expressions

After applying the Quotient Rule, we often end up with complex fractions that need simplifying. This not only makes our solutions cleaner but also easier to interpret and evaluate. With our derivative formula \( y' = \frac{(2x)(5x^6 - 1) - (x^2)(30x^5)}{(5x^6 - 1)^2} \) , the steps to simplify involved:

• Expanding both terms in the numerator, converting the product into sums of powers of \( x \).
• Combining like terms, such as \( 10x^7 \) and \( 30x^7 \), to simplify the expression.
• This gave us \( y' = \frac{-20x^7 - 2x}{(5x^6 - 1)^2} \), a much cleaner form.

Simplification is a key step to ensure your final answer is concise and easily understood.

Calculus Problem Solving

Mastering calculus requires a combination of understanding fundamental rules and applying them through problem-solving exercises. In the exercise, you first identified the components \( u \) and \( v \), differentiated using the Power Rule, and applied the Quotient Rule. Finally, you combined these results by simplifying the derivative expression. This systematic approach is crucial in solving calculus problems effectively.Here are a few key points to remember:

• Break down the problem into simple, manageable steps.
• Understand which rules apply and carefully perform the calculus operations.
• Simplify your final expression to make understanding and further calculations easier.

By consistently practicing with different functions, you'll build confidence in calculus problem-solving and become more proficient over time.