Problem 240 - Calculus Volume 1

Question

Let \(y=(f(u)+3 x)^{2}\) and \(u=x^{3}-2 x .\) If \(f(4)=6\) and \(\frac{d y}{d x}=18\) when \(x=2,\) find \(f^{\prime}(4)\)

Step-by-Step Solution

Answer

\(f'(4) = -\frac{9}{40}\).

Step 1: Differentiate Implicitly

Given the function \(y = (f(u) + 3x)^2\), we will differentiate both sides with respect to \(x\) using the chain rule. The derivative is \(\frac{dy}{dx} = 2(f(u) + 3x)\cdot (f'(u)\cdot \frac{du}{dx} + 3)\).

Step 2: Differentiate u with respect to x

Given \(u = x^3 - 2x\), differentiate \(u\) with respect to \(x\). We get \(\frac{du}{dx} = 3x^2 - 2\).

Step 3: Substitute Known Values

We know \(x = 2\), \(f(4) = 6\), and \(\frac{dy}{dx} = 18\). First, calculate \(u\) when \(x = 2\): \(u = (2)^3 - 2 \times 2 = 4\). Then substitute back into the derivative: \(18 = 2(f(4) + 3 \times 2)(f'(4) \cdot (3 \times 2^2 - 2) + 3)\).

Step 4: Simplify and Solve for f'(4)

First calculate \(f(4) + 6 = 6 + 6 = 12\). The equation for \(\frac{dy}{dx}\) becomes \(18 = 2 \times 12 \times (f'(4) \times 10 + 3)\). Simplify and solve: \(18 = 24(10 f'(4) + 3)\). Dividing both sides by 24 gives \(\frac{3}{4} = 10 f'(4) + 3\).

Step 5: Isolate f'(4)

Subtract 3 from both sides: \(\frac{3}{4} - 3 = 10 f'(4)\). Simplifying, \(-\frac{9}{4} = 10 f'(4)\). Divide by 10: \(f'(4) = -\frac{9}{40}\).

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus for finding the derivative of a composite function. When we have a function nested inside another function, like a Russian doll, the Chain Rule allows us to take the derivative of this complex structure. It states that: if a function is composed of two functions, say \( y = g(f(x)) \), then the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = g'(f(x)) \, \cdot \, f'(x) \). Here, \( g'(f(x)) \) takes care of the outer function, and \( f'(x) \) handles the inner one. This is applied by first differentiating the outer function, keeping the inner function unchanged, and then multiplying it by the derivative of the inner function. This crucial rule is especially handy for implicit differentiation, where functions aren't presented in the usual "y in terms of x" manner.

Derivative of Composite Function

Understanding how to derive a composite function, like in our example, \( y = (f(u) + 3x)^2 \), requires carefully applying the Chain Rule. A composite function, which is essentially a function inside another function, needs a step-by-step approach:

• First, focus on the outer function. Here, the outer function is \( (f(u) + 3x)^2 \), which derives to \( 2(f(u) + 3x) \). This is similar to applying the power rule.
• Next, multiply by the derivative of the inner part \( f(u) + 3x \). This entails finding derivative of both \( f(u) \) and \( 3x \).
  • For \( 3x \), the derivative is straightforward: it is 3.
  • For \( f(u) \), we chain it with \( u \) by employing \( f'(u) \cdot \frac{du}{dx} \).

This ensures each component of the composite function is correctly differentiated, leading us to the full derivative.

Implicit Differentiation Steps

Implicit differentiation is an extension of standard differentiation, used when functions are not explicitly solved for one variable. Instead of rearranging the entire expression to isolate \( y \), we differentiate both sides of the equation with respect to the intended variable, treating \( y \) as a function of \( x \).Here's how we can differentiate implicitly:

• Differentiate each term systematically, treating one variable as dependent. In our task, \( y = (f(u) + 3x)^2 \) is derived as \( \frac{dy}{dx} = 2(f(u) + 3x)(f'(u)\cdot\frac{du}{dx} + 3) \).
• Next, isolate \( \frac{dy}{dx} \) and any other unknowns, substituting known values when necessary. This is seen when plugging \( x = 2 \), \( f(4) = 6 \), and the derived equation back together.
• Finally, solve your equation for the desired derivative or variable.

Implicit differentiation simplifies complex expressions and functions and maintains their validity throughout the problem-solving process.

Function Derivatives

Derivatives offer a profound way to understand how functions change. From simple lines to complex curves, knowing how a function behaves as its arguments shift is crucial in calculus.Function derivatives, like each component of our task, involve basic differential rules:

• The Power Rule, often represented as \( \frac{d}{dx}x^n = nx^{n-1} \).
• The Sum and Difference Rule, allowing us to differentiate terms separately and simply add or subtract results.

Consider our given function \( u = x^3 - 2x \). Using the Power Rule on \( x^3 \) yields \( 3x^2 \), and the constant differentiation rule tackles \( -2x \) as \( -2 \). Together these contribute to finding \( \frac{du}{dx} = 3x^2 - 2 \).Once we have derivatives of individual functions, we can tackle composite functions, as they often require combining different rules for precise analysis.