Problem 266 - Calculus Volume 1

Question

For the following exercises, use the functions \(y=f(x)\) to find a. \(\frac{d f}{d x}\) at \(x=a\) and b. \(x=f^{-1}(y).\) c. Then use part b. to find \(\frac{d f^{-1}}{d y}\) at \(y=f(a).\) $$f(x)=9-x^{2}, 0 \leq x \leq 3, x=2$$

Step-by-Step Solution

Answer

a. \(-4\), b. \(x=2\), c. \(-\frac{1}{4}\).

Step 1: Find the Derivative of the Function

The given function is \( f(x) = 9 - x^2 \). To find \( \frac{df}{dx} \), take the derivative of \( f(x) \) with respect to \( x \). We have:\[\frac{df}{dx} = \frac{d}{dx}(9 - x^2) = 0 - 2x = -2x.\]

Step 2: Evaluate the Derivative at \(x=a\)

Now we need to evaluate \( \frac{df}{dx} \) at \( x = 2 \). Substitute \( x = 2 \) into the derivative:\[\frac{df}{dx}\bigg|_{x=2} = -2 \cdot 2 = -4.\]

Step 3: Solve for the Inverse Function

We want to find \( x = f^{-1}(y) \). Start with the equation \( y = 9 - x^2 \) and solve for \( x \) in terms of \( y \):\[y = 9 - x^2 \9 - y = x^2 \x = \sqrt{9 - y}.\]

Step 4: Evaluate the Inverse Function at \(y = f(a)\)

Find \( f(a) \) for \( a = 2 \):\[f(2) = 9 - 2^2 = 9 - 4 = 5.\]Now, find \( x = f^{-1}(5) \):\[x = \sqrt{9 - 5} = \sqrt{4} = 2.\]

Step 5: Differentiate the Inverse Function

We need the derivative of \( x = \sqrt{9 - y} \). Using the chain rule:\[x = (9 - y)^{0.5} \Rightarrow \frac{dx}{dy} = 0.5(9-y)^{-0.5}(-1) = -\frac{1}{2\sqrt{9-y}}.\]

Step 6: Evaluate the Derivative of the Inverse Function at \(y=f(a)\)

Evaluate \( \frac{df^{-1}}{dy} \) at \( y = f(2) = 5 \):\[\frac{df^{-1}}{dy}\bigg|_{y=5} = -\frac{1}{2\sqrt{9-5}} = -\frac{1}{2\cdot2} = -\frac{1}{4}.\]

Key Concepts

Derivatives

A derivative represents how a function changes as its input changes. In mathematical terms, the derivative of a function, denoted as \( \frac{df}{dx} \), measures the rate at which \( f(x) \) transforms with respect to changes in \( x \). For the function \( f(x) = 9 - x^2 \), finding the derivative involves differentiating each component of the equation. The constant 9 becomes zero, and the derivative of \( -x^2 \) is \( -2x \). Thus, \( \frac{df}{dx} = -2x \). This tells us the rate of change of \( f(x) \) is dependent on the value of \( x \).

Derivatives are foundational in calculus since they provide insight into the behavior of functions. Specifically, they enable us to assess the slope of a function's graph at any given point. When we evaluate at a specific \( x \), like \( x=2 \), we substitute into \( -2x \) to find the slope at that point, resulting in \( \frac{df}{dx}\big|_{x=2} = -4 \). This calculation reveals that at \( x=2 \), our function is decreasing at a rate of 4 units per unit increment in \( x \).

Inverse Functions

Inverse functions 'undo' the work of the original function. If a function \( f \) maps \( x \) to \( y \), the inverse function \( f^{-1} \) takes \( y \) back to \( x \). The concept is that for each output \( y \), the inverse function finds the corresponding input \( x \) that would produce \( y \).

In the given exercise, we start with \( y = 9 - x^2 \) to express \( x \) in terms of \( y \). By rearranging this equation, we find \( x = \sqrt{9 - y} \). This equation essentially tells us how to calculate \( x \) if we know \( y \). Inverting functions is particularly useful for solving equations and finding original inputs from known outputs.

For this function, we check \( f(2) \) which results in \( y=5 \), so \( f^{-1}(5)=2 \). This confirms that 2 is the initial point which, when input into \( f(x) \), gives 5.

Chain Rule

The chain rule is a fundamental principle used in calculus to differentiate composite functions. A composite function involves a function within another function. In simpler terms, if you have a function \( u(y) \) nested within another function \( g(u) \), the chain rule helps differentiate the outer function (\( g \)) with respect to the inner function (\( u \)), then multiply it by the derivative of the inner function.

The chain rule's formula can be expressed as: \( \frac{dg}{dy} = \frac{dg}{du} \cdot \frac{du}{dy} \).

In the context of the inverse function \( x = \sqrt{9 - y} \) from the original exercise, we can apply the chain rule to find the derivative. Here, the inner function is \( u(y) = 9 - y \) and the outer function is \( v(u) = u^{0.5} \). Differentiating \( v \) results in \( 0.5u^{-0.5} \) and \( u \) yields \( -1 \). By the chain rule, \( \frac{dx}{dy} = 0.5 \, \text{(-1)} \times (9 - y)^{-0.5} \). This approach helps us uncover how changes in \( y \) affect \( x \).

Evaluating Derivatives

Evaluating derivatives involves calculating the derivative of a function at a specific point. It essentially determines the slope of the tangent line to a function at a given point, offering insights into the instantaneous rate of change of the function.

Consider finding \( \frac{df}{dx} \) or \( \frac{df^{-1}}{dy} \) in our exercise. Remember, for \( f(x) = 9 - x^2 \), evaluating \( \frac{df}{dx} \) at \( x=2 \) yields \( -4 \), illustrating a slope of \(-4\) at that point. Similarly, when evaluating \( \frac{df^{-1}}{dy} \) for the inverse, it involves computing the derivative of the inverse at \( y=f(2)=5 \).

The expression \( \frac{dx}{dy} = -\frac{1}{2\sqrt{9-y}} \), evaluated at \( y=5 \), gives \( -\frac{1}{4} \). This result indicates how \( x \) changes with \( y \) in the inverse function. Recognizing these calculations helps understand how differently shaped functions behave locally at particular points.