Problem 46 - 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 \(h^{\prime}(1) .\) \(h(x)=\left(x^{2}+1\right) g(x)\)

Step-by-Step Solution

Answer

The derivative of the function \(h(x) = (x^2 + 1)g(x)\) is given by the product rule: \(h'(x) = (2x) * g(x) + (x^2 + 1) * g'(x)\). Substituting the given values at x=1, we find that \(h'(1) = (2*(-2)) + (2 * 3) = -4 + 6 = 2\).

Step 1: Apply Product Rule on h(x)

To find the derivative of h(x) = (x^2 + 1)g(x), we can use the product rule, which states: For functions u(x) and v(x), the derivative of the product u(x)v(x) is given by: (uv)' = u'v + uv' In this case, our functions are: u(x) = x^2 + 1 v(x) = g(x) Now, we need to find their derivatives: u'(x) = d(x^2+1)/dx = 2x v'(x) = g'(x) Using the product rule, the derivative of h(x) is: h'(x) = (2x) * g(x) + (x^2 + 1) * g'(x)

Step 2: Substitute the Given Values

Now, we need to find the value of h'(1) by substituting x = 1: h'(1) = (2 * 1) * g(1) + (1^2 + 1) * g'(1) Now, we will use the given values: g(1) = -2 and g'(1) = 3: h'(1) = (2*(-2)) + (2 * 3)

Step 3: Calculate h'(1)

Finally, we will calculate the value of h'(1): h'(1) = -4 + 6 = 2 Thus, the value of h'(1) is 2.

Key Concepts

Differentiation

Differentiation is like a magic tool in calculus. It helps us understand how functions change. We often want to know the rate at which something happens. For example, how fast a car is moving at a certain moment. Differentiation gives us that rate. When we differentiate a function, we find its derivative. This process involves a few rules, just like in math games.

One important rule is the Product Rule. It helps when we deal with functions multiplied together, like in the exercise above. By breaking down functions into simpler parts, the Product Rule allows us to find their derivatives more easily. This is useful in many real-world applications, where combined factors affect change. Differentiation helps us make sense of these complex interactions.

Derivatives

A derivative is like the heartbeat of a function. It tells us the speed or rate of change at any point. Think of it as a speedometer for functions. For any function you have, its derivative will give you an insight into how the function behaves. This is crucial in physics to model motion, in economics to calculate cost changes, and in engineering to predict behavior over time.

When we find the derivative of a function, we're really seeing how sensitive the function is to changes in its input. For example, if we're looking at \( h(x) = (x^2 + 1) g(x) \), the derivative, \( h'(x) \), shows us how this whole expression changes as x changes a tiny bit. This information is critical for making predictions and understanding trends.

Step-by-Step Calculus Solutions

Solving calculus problems step-by-step is like following a treasure map. Each step brings you closer to the solution with clarity. Following each rule and operation methodically ensures no steps are missed and mistakes are minimized. This methodical approach is especially important in calculus, where skipping a step can lead to incorrect answers.

• First, identify the functions involved and the operations you need, like using the Product Rule for multiplication.
• Next, substitute given values and simplify the expressions by hand.
• Finally, complete the calculation, checking your work as you go.

Through this exercise, by using the Product Rule and substituting known values, you can see how to systematically find \( h'(x) \). The step-by-step approach teaches precision, a vital skill for any mathematical problem-solving.