Problem 1 - Calculus Early Transcendentals

Question

Given that $$ f^{\prime}(0)=2, g(0)=0 \quad \text { and } \quad g^{\prime}(0)=3 $$ find \((f \circ g)^{\prime}(0)\)

Step-by-Step Solution

Answer

The derivative \((f \circ g)'(0)\) is 6.

Step 1: Understanding the Problem

We need to find the derivative of the composite function \((f \circ g)(x) = f(g(x))\) at \(x = 0\). We are given three values: \(f'(0) = 2\), \(g(0) = 0\), and \(g'(0) = 3\).

Step 2: Use the Chain Rule

The chain rule for the derivative of a composite function \((f \circ g)(x)\) states that \((f \circ g)'(x) = f'(g(x)) \cdot g'(x)\). Our task is to find \((f \circ g)'(0)\).

Step 3: Evaluate g at 0

Substitute \(x = 0\) into \(g(x)\): \(g(0) = 0\). We need this value to plug into the derivative of \(f\).

Step 4: Apply the Chain Rule at x = 0

Using the chain rule \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \), compute it at \(x = 0\):\[(f \circ g)'(0) = f'(g(0)) \cdot g'(0)\].

Step 5: Substitute Known Values

Using the given values, substitute into the chain rule expression:- \(g(0) = 0\), thus \(f'(g(0)) = f'(0) = 2\).- \(g'(0) = 3\).So, \((f \circ g)'(0) = 2 \cdot 3 = 6\).

Step 6: Final Answer

The derivative of the composite function \((f \circ g)(x)\) at \(x = 0\) is \(6\).

Key Concepts

Derivative of Composite Functions

When we talk about the derivative of composite functions, we are unraveling a process to understand how a change in one function affects another. Think of composite functions as a machine, where one function outputs a result that becomes an input for another function.
In this context, if you have a composite function like \((f \circ g)(x) = f(g(x))\), determine the derivative by seeing how small changes in \(x\) initiate changes in \(g(x)\) and consequently in \(f(g(x))\). It is important to note the interconnected dynamics at play here.
Let’s picture it with a stepwise flow to simplify:

• First, the input \(x\) changes, prompting a change in \(g(x)\).
• Then, that new \(g(x)\) value changes \(f(g(x))\).

By finding this derivative, we understand the rate of change of \(f(g(x))\) with respect to the initial variable \(x\). Understanding this mechanism will prepare you for more advanced calculus concepts.

Calculus Derivatives

Calculus derivatives serve as the foundational blocks in calculus, helping us understand how functions change. Essentially, a derivative provides us with the rate at which a function's value changes with respect to changes in its input. It’s like measuring how quickly you’re moving if you envision driving along a curvy road.
For differentiating a function \(f(x)\), the derivative is represented as \(f'(x)\) or \(\frac{df}{dx}\). This tells us how \(f\) changes as \(x\) changes infinitesimally. The objective is to find out: How does a tiny change in input (\(x\)) result in a change in the output (\(f(x)\))?
In practice, derivatives help solve various real-world problems, from calculating the speed of a vehicle to predicting future trends in economics. Their applications are vast and crucial to analytical problem-solving. As you delve further into calculus, understanding the derivation process will sharpen your analytical skills.

Chain Rule Application

The chain rule is a powerful tool in calculus, providing a method to find the derivative of composite functions. Think of it as a way to unravel the intricacy when one function is composed inside another. This rule helps when functions are nested, meaning you have \(f(g(x))\) or some variation thereof.
The chain rule formula is expressed as: \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \). To apply it effectively, follow these straightforward guidelines:

• Identify an inner function (e.g., \(g(x)\)) and an outer function (e.g., \(f(u)\) where \(u = g(x)\)).
• Differentiate the outer function with respect to the inner function: \(f'(g(x))\).
• Differentiate the inner function: \(g'(x)\).
• Multiply these derivatives together.

Applying this, you can solve our initial problem by calculating \((f \circ g)'(0) = f'(g(0)) \cdot g'(0)\). Once you substitute the known values into the chain rule expression, you understand how the combination of derivatives operates, giving insights into how tightly interconnected these functions are. Mastering this rule is a stepping stone to solving complex calculus problems with ease.