Problem 39 - Calculus and its applications

Question

Differentiate each function. \(y=\frac{\sqrt[3]{x}-7}{\sqrt{x}+3}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{-\frac{1}{6}x^{-1/6} + x^{-2/3} + \frac{7}{2}x^{-1/2}}{(x^{1/2} + 3)^2} \)."

Step 1: Rewrite function for better differentiation

Separate the numerator and denominator of the function to express it in a form suitable for using the quotient rule. Rewrite the function as: \[ y = \frac{x^{1/3} - 7}{x^{1/2} + 3} \] This allows us to use the quotient rule for differentiation.

Step 2: Apply the Quotient Rule

The quotient rule for differentiation states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \). Identify \( u = x^{1/3} - 7 \) and \( v = x^{1/2} + 3 \). Differentiate both: - \( u' = \frac{d}{dx}(x^{1/3} - 7) = \frac{1}{3}x^{-2/3} \) - \( v' = \frac{d}{dx}(x^{1/2} + 3) = \frac{1}{2}x^{-1/2} \).

Step 3: Substitute derivatives into the Quotient Rule

Substitute \( u, u', v, \) and \( v' \) into the quotient rule formula:\[ \frac{dy}{dx} = \frac{\left(\frac{1}{3}x^{-2/3}(x^{1/2} + 3) - (x^{1/3} - 7)\left(\frac{1}{2}x^{-1/2}\right)\right)}{(x^{1/2} + 3)^2} \]

Step 4: Simplify the expression

Expand and simplify the numerator:- For the first term: \( \frac{1}{3}x^{-2/3}x^{1/2} + \frac{1}{3}x^{-2/3}\cdot3 = \frac{1}{3}x^{-1/6} + x^{-2/3} \)- For the second term: \( \frac{1}{2}x^{-1/2}(x^{1/3} - 7) = \frac{1}{2}x^{-1/6} - \frac{7}{2}x^{-1/2} \)Combining these:\[ \frac{dy}{dx} = \frac{\left( \frac{1}{3}x^{-1/6} + x^{-2/3} - \frac{1}{2}x^{-1/6} + \frac{7}{2}x^{-1/2} \right)}{(x^{1/2} + 3)^2} \]

Step 5: Combine like terms

Combine the like terms in the numerator:\[ \frac{dy}{dx} = \frac{\left( \left( \frac{1}{3} - \frac{1}{2} \right)x^{-1/6} + x^{-2/3} + \frac{7}{2}x^{-1/2} \right)}{(x^{1/2} + 3)^2} \]Upon simplifying, \( \frac{1}{3} - \frac{1}{2} = -\frac{1}{6} \), so:\[ \frac{dy}{dx} = \frac{\left(-\frac{1}{6}x^{-1/6} + x^{-2/3} + \frac{7}{2}x^{-1/2}\right)}{(x^{1/2} + 3)^2} \]

Key Concepts

Understanding the Quotient Rule

When tackling problems involving calculus differentiation, one fundamental tool at our disposal is the quotient rule. It's specifically used when we have a function that is the quotient or division of two other functions. Simply put, if we need to find the derivative of a function defined as \( f(x) = \frac{u(x)}{v(x)} \), where \( u(x) \) and \( v(x) \) are differentiable functions, we apply the quotient rule.
The formula for the quotient rule is:

• If \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \)

To break it down:

• \( u' \) is the derivative of the numerator \( u \).
• \( v' \) is the derivative of the denominator \( v \).
• We multiply \( u' \) by \( v \) and subtract \( u \) multiplied by \( v' \).
• Finally, we divide the entire expression by the square of \( v \).

This rule is advantageous because it helps maintain consistency and accuracy when differentiating quotient functions. Not applying this properly can lead to unwanted errors in the result.
So, always remember: when a problem involves the division of two functions, the quotient rule is your go-to strategy!

Discovering Derivatives

Derivatives are a core concept in calculus, representing the rate of change of a function with respect to its variable, often time or space. In a more graphical sense, if you have a curve on a graph, the derivative at any given point gives us the slope of the tangent line to that curve at that point.
Understanding derivatives begins with recognizing basic differentiation rules:

• The derivative of \( x^n \) (where \( n \) is any real number) is \( nx^{n-1} \).
• The derivative of a constant is always zero.
• The derivative of the sum of two functions is the sum of their derivatives.

Let's go through an example: Differentiating the function \( u(x) = x^{1/3} - 7 \) involves using the power rule, resulting in a derivative \( u'(x) = \frac{1}{3}x^{-2/3} \). Similarly, for \( v(x) = x^{1/2} + 3 \), applying the power rule gives us \( v'(x) = \frac{1}{2}x^{-1/2} \). Understanding this process allows us to break down complex expressions into simpler parts, making it easier to apply rules like the quotient rule in calculus differentiation.

The Art of Function Simplification

Function simplification is a valuable step in calculus problems, especially when preparing to apply differentiation techniques like the quotient rule. Simplifying a function involves reformatting it in a way that makes the application of these rules easier and straightforward.
Consider our original function: \( y = \frac{\sqrt[3]{x} - 7}{\sqrt{x} + 3} \). Attempting to differentiate this directly can be daunting. To simplify, we rewrite the roots as exponents: \( \sqrt[3]{x} \) becomes \( x^{1/3} \), and \( \sqrt{x} \) becomes \( x^{1/2} \). Our function then morphs into: \( y = \frac{x^{1/3} - 7}{x^{1/2} + 3} \).
This transformation is crucial for several reasons:

• It eases the differentiation process by enabling the straightforward application of the power rule.
• Exponent notation is often more manageable algebraically, especially when calculating derivatives.
• It helps in clearly defining the numerator \( u \) and the denominator \( v \) for quick application of the quotient rule.

By simplifying functions, we set ourselves up for success in applying calculus differentiation effectively!