Problem 77 - Calculus for Biology and Medicine

Question

Differentiate $$ y=\frac{e^{2 x}(9 x-2)^{3}}{\sqrt[4]{\left(x^{2}+1\right)\left(3 x^{3}-7\right)}}. $$

Step-by-Step Solution

Answer

Use the quotient rule, differentiate and simplify: \(\frac{dy}{dx} = \frac{v(du/dx) - u(dv/dx)}{v^2}\).

Step 1: Identify the Differentiation Rule

To differentiate the function \(y = \frac{e^{2x} (9x-2)^3}{\sqrt[4]{(x^2 + 1)(3x^3 - 7)}}\), we notice that it is a quotient. We'll use the quotient rule, which states \(\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\), where \(u = e^{2x}(9x-2)^3\) and \(v = ((x^2 + 1)(3x^3 - 7))^{1/4}\).

Step 2: Differentiate the Numerator

Differentiate \(u = e^{2x}(9x - 2)^3\) using the product rule. Let \(f(x) = e^{2x}\) and \(g(x) = (9x - 2)^3\). Then, \(\frac{du}{dx} = \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)\).Calculate:- \(f'(x) = 2e^{2x}\).- \(g'(x) = 3(9x - 2)^2 \cdot 9 = 27(9x - 2)^2\).Thus, \(\frac{du}{dx} = 2e^{2x}(9x - 2)^3 + e^{2x} \cdot 27(9x - 2)^2\).

Step 3: Differentiate the Denominator

Find \(\frac{dv}{dx}\) for \(v = ((x^2 + 1)(3x^3 - 7))^{1/4}\) using the chain rule and the product rule. Let \(h(x) = (x^2 + 1)\) and \(k(x) = (3x^3 - 7)\).First, define \((x^2 + 1)(3x^3 - 7) = h(x)k(x)\) then the derivative is given by:- \(h'(x) = 2x\).- \(k'(x) = 9x^2\).- \(\frac{d}{dx}[h(x)k(x)] = (2x)(3x^3 - 7) + (x^2 + 1)(9x^2)\).Then, apply the chain rule, with \(v' = \frac{1}{4}((x^2 + 1)(3x^3 - 7))^{-3/4} \cdot \frac{d}{dx}[h(x)k(x)]\).

Step 4: Apply the Quotient Rule

Now use the results of differentiated \(u\) and \(v\) with the quotient rule: \[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]Substitute the values and simplify the expression for \(\frac{dy}{dx}\), ensuring each calculated derivative is correctly used in the formula.

Step 5: Simplify the Expression

Combine and simplify all terms in the expression for \(\frac{dy}{dx}\). This often involves algebraic manipulation like expanding brackets, combining like terms, and simplifying powers. Ensure that the simplified version corresponds to the correct derivation processes for both numerator and denominator terms under the quotient rule.

Step 6: Verify the Solution

Double-check each derivative computation and simplification. Make sure that any algebraic errors are corrected and that the resulting expression for \(\frac{dy}{dx}\) is properly simplified and follows logically from applying the quotient rule.

Key Concepts

Quotient Rule

When faced with differentiating a function that appears as a fraction or a quotient, you need to use the quotient rule. This rule is specifically designed for scenarios where functions are divided by each other. For the function\[ y = \frac{e^{2x} (9x-2)^3}{\sqrt[4]{(x^2 + 1)(3x^3 - 7)}} \]we identify that there's a clear division between the numerator \( u = e^{2x}(9x-2)^3 \) and the denominator \( v = ((x^2 + 1)(3x^3 - 7))^{1/4} \). The quotient rule formula is:\[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]To apply it, follow these steps:

• Differentiate the numerator \( \frac{du}{dx} \)
• Differentiate the denominator \( \frac{dv}{dx} \)
• Substitute these derivatives back into the quotient rule formula
• Simplify your expression for clarity

This method helps ensure that the differentiation respects the division relationship between the two functions.

Product Rule

Often, when dealing with differentiation, you encounter functions that are the product of two or more simpler functions. In such cases, employing the product rule is crucial. This rule is specifically crafted for differentiating products of functions. Given the numerator in our exercise, we have\[ u = e^{2x}(9x-2)^3 \]Here, the product rule allows us to differentiate by treating one function as \( f(x) = e^{2x} \) and another as \( g(x) = (9x-2)^3 \). The formula for the product rule is:\[ \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \]To use it:

• Find the derivative of \( f(x) \), noted as \( f'(x) = 2e^{2x} \)
• Find the derivative of \( g(x) \) noted as \( g'(x) = 3(9x-2)^2 \cdot 9 = 27(9x-2)^2 \)
• Plug these derivatives into the product rule formula
• Combine like terms to simplify the expression

Through these steps, you're able to effectively tackle products within the differentiation process.

Chain Rule

The chain rule is a powerful tool in differentiation, especially useful when you're dealing with composite functions. A composite function is when one function is nested inside another, such as our denominator:\[ v = ((x^2 + 1)(3x^3 - 7))^{1/4} \]To differentiate this, you must first recognize it as a chain of functions, involving both a product and a power. The chain rule formula is:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]Steps to apply:

• Consider \( f(x) = x^{1/4} \) and \( g(x) = (x^2 + 1)(3x^3 - 7) \)
• Differentiating \( g(x) \) involves the product rule: first find \( h'(x) = 2x \) and \( k'(x) = 9x^2 \) then combine as per the product rule
• Apply the derivative of the outer function \( f \), resulting in \( \frac{1}{4}g(x)^{-3/4} \)
• Combine the derivatives using the chain rule

This systematic approach allows the handling of complex functions that seem tangled within one another, simplifying the differentiation process systematically.