Problem 45 - Single Variable Calculus

Question

Suppose that \(f\) and \(g\) are functions that are differentiable at \(x=1\) and that \(f(1)=2, f^{\prime}(1)=-1\), \(g(1)=-2\), and \(g^{\prime}(1)=3 .\) Find \(h^{\prime}(1) .\) . \(h(x)=f(x) g(x)\)

Step-by-Step Solution

Answer

To find the value of \(h'(1)\), we use the product rule \(h'(x) = f'(x)g(x) + f(x)g'(x)\) and substitute the given values: \(h'(1) = f'(1)g(1) + f(1)g'(1) = (-1)(-2) + (2)(3) = 2 + 6 = 8\). Therefore, \(h'(1) = 8\).

Step 1: Recall the product rule for differentiation

The product rule states that if \(u(x)\) and \(v(x)\) are differentiable functions, then the derivative of their product is given by: \[ (uv)'(x) = u'(x)v(x) + u(x)v'(x) \]

Step 2: Apply the product rule to \(h(x)\)

In this case, \(u(x) = f(x)\) and \(v(x) = g(x)\), so the derivative of \(h(x)\) can be calculated as: \[ h'(x) = f'(x)g(x) + f(x)g'(x) \]

Step 3: Evaluate the derivative at \(x = 1\)

Substitute the given values of \(f(1)\), \(f'(1)\), \(g(1)\), and \(g'(1)\) into the expression for \(h'(1)\): \[ h'(1) = f'(1)g(1) + f(1)g'(1) \] Using the given information, we have: \[ h'(1) = (-1)(-2) + (2)(3) = 2 + 6 \]

Step 4: Find the value of \(h'(1)\)

Add the values obtained in the previous step to get the result: \[ h'(1) = 2 + 6 = 8 \] Thus, the value of \(h'(1)\) is 8.

Key Concepts

Differentiable Functions

Before diving into derivative calculations, it's important to understand differentiable functions. A function is differentiable at a point if it has a derivative at that point. In mathematical terms, if a function is differentiable, it means you can calculate a specific rate at which the function's value is changing at any given point.
Differentiable functions are smooth, meaning they don't have any sharp corners or discontinuities. Imagine them as a smooth track where the direction changes gently. In our exercise, both functions \( f(x) \) and \( g(x) \) are differentiable at \( x=1 \). This sets the stage for us to use the product rule in the next section.

Derivative Calculation

Once we confirm that our functions are differentiable, we can move on to calculating derivatives. Derivatives measure how a function's output changes as its input changes. This is crucial for understanding dynamics in fields like physics, economics, and biology.
In our specific exercise, we need the derivative of \( h(x) = f(x)g(x) \). Here, we'll utilize the famous product rule, which is a technique to find the derivative of a product of two functions. According to the product rule:

• If \( u(x) \) and \( v(x) \) are two differentiable functions, then their product \((uv)(x)\) has a derivative given by:
\[ (uv)'(x) = u'(x) v(x) + u(x) v'(x) \]

This rule allows us to break down the problem and apply the derivatives calculated individually to find \( h'(x) \).

Evaluation of Derivatives

With derivatives calculated step-by-step, evaluating them at specific points gives us valuable information, like the slope or rate of change at that point. In our exercise, we need to evaluate \( h'(x) \) at \( x=1 \).
Using the product rule, we've established that:\[ h'(x) = f'(x)g(x) + f(x)g'(x) \]Now at \( x=1 \), we substitute the given values of \( f(1) \), \( f'(1) \), \( g(1) \), and \( g'(1) \) into the equation to find \( h'(1) \):

• \( h'(1) = (-1)(-2) + (2)(3) \)
• Simplifying gives \( h'(1) = 2 + 6 = 8 \)

Thus, \( h'(1) = 8 \), so at \( x=1 \), the rate at which \( h(x) \) changes is 8. This entire process demonstrates the power of calculus in understanding and evaluating the behaviors of functions at specific points.