Problem 15 - Calculus and its applications

Question

Differentiate each function $$ y=\frac{4 x^{2}}{(7-5 x)^{3}} $$

Step-by-Step Solution

Answer

Use the quotient and chain rules to differentiate: \[ y' = \frac{8x(7-5x)^3 + 60x^2(7-5x)^2}{(7-5x)^6} \]

Step 1: Identify the Differentiation Technique Needed

The function given is a quotient of two functions, which means we need to use the quotient rule for differentiation. Recall the quotient rule: if we have a function \( y = \frac{u(x)}{v(x)} \), its derivative is \( y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).

Step 2: Define the Numerator and Denominator Functions

Let \( u(x) = 4x^2 \) and \( v(x) = (7-5x)^3 \). Next, we need to find the derivatives \( u'(x) \) and \( v'(x) \).

Step 3: Differentiate the Numerator Function

Differentiate \( u(x) = 4x^2 \). Using the power rule, we find \( u'(x) = 8x \).

Step 4: Differentiate the Denominator Function

Differentiate \( v(x) = (7-5x)^3 \). Use the chain rule: let \( z = 7 - 5x \), then \( v(x) = z^3 \). Thus, \( v'(x) = 3z^2 (-5) = -15(7-5x)^2 \).

Step 5: Apply the Quotient Rule to Find the Derivative

Plug \( u(x) \), \( v(x) \), \( u'(x) \), and \( v'(x) \) into the quotient rule: \[y' = \frac{(8x)(7-5x)^3 - (4x^2)(-15)(7-5x)^2}{((7-5x)^3)^2}\]

Step 6: Simplify the Expression for the Derivative

The simplified form of the derivative is \[y' = \frac{8x(7-5x)^3 + 60x^2(7-5x)^2}{(7-5x)^6}\]Factoring out common terms and simplifying further will lead us to the final expression.

Key Concepts

Quotient Rule

When you're dealing with differentiating functions that are divided by each other, you use what's called the Quotient Rule. This rule can be remembered from the formula \( y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \), where \( u(x) \) and \( v(x) \) are the numerator and the denominator functions respectively.
Here's how it works in practice:

• Start by identifying your numerator \( u(x) \) and your denominator \( v(x) \).
• Differentiating \( u(x) \) gives you \( u'(x) \).
• Differentiating \( v(x) \) gives you \( v'(x) \).
• Substitute these derivatives into the quotient rule formula.

The Quotient Rule is especially helpful when you can't simplify a function to avoid a division, making it indispensable for calculus. Use it whenever a fraction is present in a given function, like in \( y = \frac{4x^2}{(7 - 5x)^3} \). In this case, \( u(x) = 4x^2 \) and \( v(x) = (7 - 5x)^3 \). These steps will help you efficiently derive the function.

Power Rule

The Power Rule is a basic rule that is easy to use. Whenever you differentiate terms like \( x^n \), you can apply this rule. It's simply expressed as \( \frac{d}{dx}[x^n] = nx^{n-1} \).
To use it effectively, follow these steps:

• Identify all terms that look like powers, such as \( x^2 \) in the numerator function \( u(x) = 4x^2 \).
• Apply the rule by multiplying the current power by the coefficient (e.g., \( 2 \times 4 = 8 \) for \( 4x^2 \)).
• Reduce the power by one (turn \( x^2 \) into \( x^{1} \)).

This straightforward method allows you to differentiate polynomials quickly, as seen when finding \( u'(x) = 8x \). It's one of the first and most intuitive differentiation methods students should master.

Chain Rule

The Chain Rule is essential for differentiating composite functions, where you have a function within another function. It's useful in expressions like \( v(x) = (7 - 5x)^3 \). Consider this expression as \( f(g(x)) \), where \( g(x) = 7 - 5x \) and \( f(z) = z^3 \). The rule is represented by \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).
Here's a simple way to apply the Chain Rule:

• Identify the inner function \( g(x) \) and the outer function \( f(z) \), where \( z = g(x) \).
• Differentiating the outer function: find \( f'(z) \) (like \( 3z^2 \) from \( z^3 \)).
• Differentiating the inner function: derive \( g(x) \). For \( g(x) = 7 - 5x \), differentiation gives \( -5 \).
• Multiply the derivatives: obtain \( g'(x) \cdot f'(g(x)) = (-5) \times 3(7 - 5x)^2 = -15(7-5x)^2 \).

Mastering the Chain Rule is crucial, especially in compound expressions where simple differentiation rules don't apply directly. Through these methods, students can confidently tackle a wide range of problems.