Problem 240 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The derivative \(f'(4)\) is \(-\frac{9}{40}\).

Step 1: Understand the Problem

We need to find the derivative of the function given as \(y = (f(u) + 3x)^2\) with respect to \(x\), where \(u = x^3 - 2x\). We know that \(f(4) = 6\) and \(\frac{dy}{dx} = 18\) when \(x = 2\). Our goal is to find \(f'(4)\).

Step 2: Differentiate the Outer Function

Using the chain rule, differentiate \(y = (f(u) + 3x)^2\) with respect to \(x\). Let's denote \(v = f(u) + 3x\). Then, \(y = v^2\). The derivative of \(y\) with respect to \(v\) is \(\frac{dy}{dv} = 2v\).

Step 3: Differentiate the Inner Function

We need the derivative of \(v = f(u) + 3x\) with respect to \(x\). Differentiating, we get \(\frac{dv}{dx} = f'(u)\frac{du}{dx} + 3\).

Step 4: Differentiate with Respect to \(x\)

Now differentiate \(u = x^3 - 2x\), which gives us \(\frac{du}{dx} = 3x^2 - 2\). Therefore, \(\frac{dv}{dx} = f'(u)(3x^2 - 2) + 3\).

Step 5: Apply the Chain Rule

Using the chain rule, \(\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx} = 2v \cdot (f'(u)(3x^2 - 2) + 3)\).

Step 6: Solve for \(x = 2\)

Substitute \(x = 2\) into the equations. First, calculate \(u = 2^3 - 2\times2 = 8 - 4 = 4\). Therefore \(f(u) = f(4) = 6\) and \(v = f(u) + 3x = 6 + 3\times2 = 12\).

Step 7: Substitute Known Values

We know \(\frac{dy}{dx} = 18\) when \(x = 2\). Substitute into the equation: \(18 = 2(12)(f'(4)(3\times2^2 - 2) + 3)\). Simplify to get \(18 = 24(f'(4) \cdot 10 + 3)\), which simplifies to: \(18 = 240f'(4) + 72\).

Step 8: Solve for \(f'(4)\)

Subtract 72 from both sides, we get: \(-54 = 240f'(4)\). Now solve for \(f'(4)\): \(f'(4) = -\frac{54}{240} = -\frac{9}{40}\).

Key Concepts

Derivatives

Derivatives are such a crucial part of calculus. Essentially, they tell us about how a function changes at any specific point. Think of it as the speed of a car at any moment on your drive. In mathematical terms, it's the slope of the curve at any point.

• If you've got a straight line, the derivative is constant.
• If your function is curved, the derivative changes as you move along.

In our exercise, we are given a derivative of a function with respect to another variable. We're tasked with finding how quickly one function changes when another variable is tweaked, using derivatives. This is done by computing derivative formulas and using them to find unknowns as required.

Chain Rule

The chain rule is like a toolbox for calculus that helps when dealing with complex functions made up of other functions. In simpler terms, when you have a function nested within another, like peeling an onion, the chain rule lets you peel each layer.

• You differentiate the outer function first.
• Then multiply by the derivative of the inner function.

This is crucial when you have composite functions, like \( y = (f(u) + 3x)^2 \). Here, we apply the chain rule by differentiating the outer layer and then the inner part. This helps link the derivative of complex functions in steps, simplifying our problem-solving process.

Implicit Differentiation

Implicit differentiation is a valuable tool when functions are not given explicitly. Sometimes, relationships between variables aren't straightforward, and that's where implicit differentiation helps. It allows us to compute derivatives without needing to first solve the equation for one particular variable.

• Take the derivative of both sides of the equation.
• Apply the chain rule when necessary.

In the given problem, implicit differentiation enables calculation of each part's derivative, such as when dealing with the inner variable \( u = x^3 - 2x \), without rearranging the equation. This method can be incredibly handy when working with mixed functions of different variables.

Function Composition

Function composition is like putting two functions together to form a new one, where the output of one function becomes the input of another.

• Consider \( f(u) \) as one function and \( u = x^3 - 2x \) as another.
• These are combined in the problem as \( y = (f(u) + 3x)^2 \).

Here, composition allows us to deal with a new function that integrates both \( f \) and \( u \), leading to a more complex relationship. Understanding function composition helps when you need to derive or analyze such nested functions, like when differentiating them using the chain rule.