Problem 40 - Calculus and its applications

Question

Differentiate each function. \(y=\frac{\sqrt{x}+4}{\sqrt[3]{x}-5}\)

Step-by-Step Solution

Answer

Differentiate using the quotient rule and simplify: \( \frac{\frac{1}{2} x^{-1/6} - \frac{1}{3} x^{-1/6} - \frac{5}{2} x^{-1/2} - \frac{4}{3} x^{-2/3}}{(x^{1/3} - 5)^2}\)."

Step 1: Identify the Function's Form

The given function is a ratio of two expressions, one involving a square root and the other a cube root. This suggests the use of the quotient rule for differentiation, which is applicable when differentiating a quotient of two functions.

Step 2: Rewrite the Function

For easier differentiation, rewrite the function as \( y = \frac{x^{1/2} + 4}{x^{1/3} - 5} \). This makes it clear how to differentiate each part using power rules.

Step 3: Apply the Quotient Rule

The quotient rule states that \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \). Here, \( u = x^{1/2} + 4 \) and \( v = x^{1/3} - 5 \). We will find the derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) next.

Step 4: Differentiate the Numerator

The numerator \( u = x^{1/2} + 4 \) differentiates to \( \frac{du}{dx} = \frac{1}{2}x^{-1/2} \). The constant 4 differentiates to zero.

Step 5: Differentiate the Denominator

The denominator \( v = x^{1/3} - 5 \) differentiates to \( \frac{dv}{dx} = \frac{1}{3}x^{-2/3} \). The constant -5 differentiates to zero.

Step 6: Plug into the Quotient Rule

Substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into the quotient rule formula. This gives:\[\frac{d}{dx}\left( \frac{x^{1/2} + 4}{x^{1/3} - 5} \right) = \frac{(x^{1/3} - 5)(\frac{1}{2}x^{-1/2}) - (x^{1/2} + 4)(\frac{1}{3}x^{-2/3})}{(x^{1/3} - 5)^2}\]

Step 7: Simplify the Expression

Simplify both the numerator and denominator. Start by multiplying through by the derivatives, then combine like terms, being mindful to simplify exponents:\[\frac{\frac{1}{2} x^{-1/2} (x^{1/3} - 5) - \frac{1}{3} x^{-2/3} (x^{1/2} + 4)}{(x^{1/3} - 5)^2}\]Careful worrking through results in:\[= \frac{\frac{1}{2} x^{-1/6} - \frac{5}{2} x^{-1/2} - \frac{1}{3} x^{-1/6} - \frac{4}{3} x^{-2/3}}{(x^{1/3} - 5)^2}\]

Step 8: Final Result

Combine terms where possible:\[= \frac{\frac{1}{2} x^{-1/6} - \frac{1}{3} x^{-1/6} - \frac{5}{2} x^{-1/2} - \frac{4}{3} x^{-2/3}}{(x^{1/3} - 5)^2}\]This is the simplified derivative of the original function.

Key Concepts

Quotient Rule

When dealing with a function that is a ratio of two differentiable functions, the quotient rule is a key tool for finding the derivative. Suppose you have a function expressed as \( \frac{u}{v} \), where \( u \) and \( v \) are both differentiable functions of \( x \). The quotient rule tells us how to find the derivative of this ratio:

• First, differentiate the numerator, \( u \), with respect to \( x \), denoted as \( \frac{du}{dx} \).
• Then, differentiate the denominator, \( v \), with respect to \( x \), denoted as \( \frac{dv}{dx} \).
• Apply the formula: \[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\]

This rule allows you to find the derivative of the entire function in terms of the derivatives of \( u \) and \( v \). It's crucial to correctly apply the rule, ensuring each derivative and term is calculated accurately.

Power Rule

The power rule is one of the simplest tools in calculus used for differentiation, particularly useful for functions that are polynomial-like in form. The basic formula for the power rule is very straightforward:

• If you have a function \( x^n \), where \( n \) is any real number, its derivative is \( nx^{n-1} \).
• This rule helps simplify the differentiation process by turning the exponent \( n \) into a helpful coefficient.

For example, applying the power rule to \( x^{1/2} \) yields \( \frac{1}{2}x^{-1/2} \), and applying it to \( x^{1/3} \) yields \( \frac{1}{3}x^{-2/3} \). Using this rule in combination with others allows you to differentiate more complex expressions efficiently.

Derivative Simplification

After applying rules like the quotient rule and power rule, simplifying the derivative expression is vital. Simplification often involves:

• Combining like terms, which means adding or subtracting parts of the expression that have the same exponent.
• Reducing fractions and powers to make the expression more compact and easier to understand.

For instance, in our problem, the expression for the derivative involves canceling out similar terms and making coefficients and exponents more straightforward. Proper simplification reveals the underlying structure of the equation, making it much easier to work with or analyze further.

Radical Expressions

Radical expressions involve roots, such as square roots or cube roots. These can initially seem challenging to differentiate but can be simplified into expressions that use the power rule.

• A square root of \( x \) is transformed into \( x^{1/2} \).
• A cube root of \( x \) becomes \( x^{1/3} \).

By rewriting radical expressions this way, you align them with the requirements of the power rule, making it possible to apply simplification techniques. Breaking down radicals into exponents is a valuable technique to simplify differentiation and integration tasks because it leverages the more straightforward manipulation of exponents rather than roots.