Problem 252 - Calculus Volume 1

Question

Use the information in the following table to find \(h^{\prime}(a)\) at the given value for \(a\). $$\begin{array}{|c|c|c|c|c|} \hline x & f(x) & f^{\prime}(x) & g(x) & g^{\prime}(x) \\ \hline 0 & 2 & 5 & 0 & 2 \\ \hline 1 & 1 & -2 & 3 & 0 \\ \hline 2 & 4 & 4 & 1 & -1 \\ \hline 3 & 3 & -3 & 2 & 3 \\ \hline \end{array}$$ $$ h(x)=f(g(\sin x)) ; a=0 $$

Step-by-Step Solution

Answer

The derivative \( h'(0) \) is 10.

Step 1: Understand the function

We have the composite function, \( h(x) = f(g(\sin x)) \). To find \( h'(a) \), we need to use the chain rule since \( h(x) \) is composed of multiple functions.

Step 2: Apply the Chain Rule

The chain rule for the derivative of \( h(x) = f(g(\sin x)) \) is \( h'(x) = \frac{df}{du} \cdot \frac{dg}{dv} \cdot \frac{d}{dx}(\sin x) \). Here, \( u = g(\sin x) \) and \( v = \sin x \).

Step 3: Differentiate each part

First differentiate \( \sin x \) to get \( \cos x \). Then, differentiate \( g(v) \) with respect to \( v \), yielding \( g'(v) \). Finally, differentiate \( f(u) \) with respect to \( u \), yielding \( f'(u) \).

Step 4: Find the value at \(a = 0\)

At \( x = 0 \), we calculate: - \( \sin(0) = 0 \)- \( g(0) = 0 \) (from the table)- \( g'(0) = 2 \) (from the table)- \( f(0) = 2 \) (from the table), but we are interested in \( f'(g(0)) = f'(0) = 5 \)Thus, substituting into the derivative formulas:\[ h'(0) = f'(g(0)) \cdot g'(0) \cdot \cos(0) \]

Step 5: Calculate the derivative

Using the values from the table and previous steps, we have: \[ h'(0) = 5 \cdot 2 \cdot 1 = 10 \] because \( \cos(0) = 1 \).

Key Concepts

Composite Functions

Composite functions are an essential concept in calculus, helping us manage complex expressions by combining different functions into a single one. In our exercise, the function is given by \( h(x) = f(g(\sin x)) \). This is a composition of three functions: first, the sine function \( \sin x \), second, the function \( g(x) \), and finally, the function \( f(x) \).
When working with composite functions, it’s helpful to think of each layer as proceeding step-by-step:

• First, process the input \( x \) through \( \sin x \).
• Then, take the result and apply \( g \) to it.
• Lastly, apply \( f \) to the outcome of \( g \).

Each function affects the final outcome step by step, creating an intricate but structured expression that we can analyze further with differentiation techniques.
These layers are why it is known as a "composite" function - it's composed of many interconnected parts, each contributing uniquely to the final result.

Derivative Calculations

Derivative calculations for composite functions can get tricky, which is why we use the Chain Rule. The Chain Rule helps us differentiate a composition of functions like \( h(x) = f(g(\sin x)) \). In this exercise, we apply it as follows:

• First, identify each part of the composite expression. Here, \( u = g(\sin x) \) and \( v = \sin x \).
• The derivative we seek is \( h'(x) = \frac{df}{du} \cdot \frac{dg}{dv} \cdot \frac{d}{dx}(\sin x) \).

By breaking the function down into manageable parts like this, we can systematically find each derivative step by step.
For example, start by differentiating \( \sin x \), which gives \( \cos x \). Next, differentiate \( g(v) \) with respect to \( v \), which gives \( g'(v) \). Lastly, differentiate \( f(u) \) with respect to \( u \), producing \( f'(u) \).
This methodological approach using the chain rule makes finding the derivative of composite functions feasible and structured, allowing you to solve complex derivative problems with ease.

Differentiation Techniques

Differentiation techniques are tools that assist in understanding and solving calculus problems involving rates of change and slopes of curves. Using these techniques, especially the Chain Rule, makes it possible to handle complex functions like compound functions efficiently.
In our exercise, the task is to differentiate the function \( h(x) = f(g(\sin x)) \) at \( a = 0 \). Here’s a step-by-step approach:

• Begin by identifying derivatives step by step: first \( \cos(x) \) from \( \sin(x) \), then \( g'(v) \), followed by \( f'(u) \).
• Use the given table values to find these derivatives at required points. For \( x=0 \): \( \sin(0)=0 \), \( g(0)=0 \), \( g'(0)=2 \), \( f'(g(0))=5 \).

Next, you substitute these values into the chain rule format: \[ h'(0) = f'(g(0)) \cdot g'(0) \cdot \cos(0) \], which simplifies the calculation. Finally, compute to get the ultimate derivative \( h'(0) = 5 \cdot 2 \cdot 1 = 10 \).
Learning and applying these differentiation techniques is key to mastering calculus and confidently approaching even the most complex functions.