Question

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u. \end{array}\right.$$ a. Show that \(\lim _{x \rightarrow u} H(v)=0\) b. For any value of \(u\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u).$$ c. Show that. $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right).$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\).

Short answer

Question: Show that if f is differentiable at g(a), then the derivative of h(x)=f(g(x)) is given by h'(a)=f'(g(a))g'(a). Solution: We were able to show that the limit of H(v) approaches 0 as v approaches u. We then established that f(v)-f(u)=(H(v)+f'(u))(v-u) and used this result to derive the expression h'(a) = lim (x→a) (H(g(x)) + f'(g(a))) • (g(x)-g(a))/(x-a). With the knowledge that lim (x→a) H(g(x)) = 0, we simplified the expression to h'(a) = f'(g(a)) lim (x→a) (g(x)-g(a))/(x-a). Finally, using the definition of the derivative, we demonstrated that h'(a) = f'(g(a))g'(a), proving the chain rule for the given function.

Step-by-step solution

Show that the limit of \(H(v)\) approaches \(0\) as \(v\) approaches \(u\).

To show that \(\lim_{x\rightarrow u} H(v) =0\), we need to examine the continuity of \(H(v)\) at \(u\). Notice that $$H(u)=\left\\{\begin{array}{ll} \frac{f(u)-f(u)}{u-u}-f'(u) & \text{ if } u \neq u\\ 0 & \text{ if } u = u \end{array}\right.$$, both cases give us \(H(u) = 0\). Since \(H(v)\) is defined piecewise, we consider the case when \(v \neq u\) and show that it is equal to the case when \(v = u\) in the limit. Since \(f\) is differentiable, it is also continuous, so \(\lim_{v\rightarrow u} f(v) = f(u)\). Using this property, we can find the limit as follows: $$\lim_{v\rightarrow u} H(v)= \lim_{v\rightarrow u} \frac{f(v)-f(u)}{v-u} - f'(u) = \lim_{v\rightarrow u} \frac{f(v)-f(u)}{v-u} - \lim_{x\rightarrow u} f'(u).$$ Since the limit of the derivative \(f'(u)\) is a constant, it remains the same as \(v\) approaches \(u\). Then, we can apply the definition of the derivative to the first part: $$\lim_{v\rightarrow u}\frac{f(v)-f(u)}{v-u} = f'(u).$$ Together, we find: $$\lim_{v\rightarrow u} H(v) = f'(u) - f'(u) = 0.$$

Show that \(f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u)\) for any value of \(u\).

We are given \(H(v)\), and we'd like to show that the given expression for the difference of the function values is true. From the definition of \(H(v)\), we can isolate \(f(v) - f(u)\): $$f(v)-f(u) = H(v)(v-u)+f'(u)(v-u).$$ Notice that when we factor \((v-u)\), we get \((v-u)(H(v)+f'(u))\). Thus, for any value of \(u\), we have shown that $$f(v)-f(u)=(H(v)+f^{\prime}(u))(v-u).$$

Show that \(h^{\prime}(a) = \lim_{x\rightarrow a} \left((H(g(x)) + f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\).

We know that the derivative \(h^{\prime}(a)\) can be written as follows: $$h^{\prime}(a) = \lim_{x\rightarrow a} \frac{h(x)-h(a)}{x-a},$$ where \(h(x) = f(g(x))\). Then, we can rewrite \(h'(a)\): $$\begin{aligned} h^{\prime}(a) &=\lim_{x\rightarrow a} \frac{f(g(x))-f(g(a))}{x-a} \\ &=\lim_{x\rightarrow a} \frac{\left(H(g(x))+f^{\prime}(g(a))\right)(g(x)-g(a))}{x-a} \end{aligned}.$$

Show that \(h^{\prime}(a) = f^{\prime}(g(a)) g^{\prime}(a)\).

Now, we can show that \(h'(a)=f'(g(a))g'(a)\) using the expression we derived in the previous step: $$h^{\prime}(a) = \lim_{x\rightarrow a} \left((H(g(x)) + f^{\prime}(g(a)))\cdot \frac{g(x)-g(a)}{x-a}\right).$$ Since \(H(v) = \lim_{v\rightarrow u}\frac{f(v) - f(u)}{v-u}-f'(u)\), we derived in step 1 that $$\lim_{v\rightarrow u} H(v)=0$$ which means that $$\lim_{x\rightarrow a} H(g(x))=0.$$ Thus, the original expression simplifies to: $$h^{\prime}(a) = \lim_{x\rightarrow a}\left(f^{\prime}(g(a)) \cdot \frac{g(x)-g(a)}{x-a}\right).$$ Since \(f'(g(a))\) is constant with respect to \(x\), we can use the chain rule to further simplify the expression: $$h^{\prime}(a)=f^{\prime}(g(a))\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}.$$ From the definition of the derivative, we know that $$\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a} = g^{\prime}(a).$$ Finally, we get: $$h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a).$$

Key concepts

Limits and Continuity

Understanding the behavior of functions as they approach a specific point is essential in calculus, and this is precisely where limits come into play. Limits help us define the value that a function approaches as the input approaches a particular value. For instance, when examining the function at a point where it might not be explicitly defined, limits provide a way to talk about the function's behavior near that point.

This concept is tightly interwoven with continuity, which is a property of a function if it is smooth or unbroken when graphed. Essentially, a function is continuous at a point if the following three conditions are met: the function is defined at the point, the limit exists at the point, and the function's value at that point equals the limit value. In the provided textbook exercise, we are asked to prove that the limit of a piecewise function as it approaches a particular point is zero, highlighting the function's continuity at that point. In practical terms, this means that there is no abrupt change or 'jump' in the function's value at that particular point.

The use of limits is not just confined to ensuring continuity, but they are also core to the fundamentals of calculus, including derivative calculations, which bring us to the concept of differentiable functions.

Derivatives

The derivative of a function at a point essentially gives you the slope of the tangent line to the function's graph at that point. It's a measure of how the function's output value changes as its input value changes. In more familiar terms, you can think of derivatives as a way to find out the 'rate of change' or to measure 'instantaneous speed' for variable rates.

In the problem provided, we explore the derivative of a composite function using an assistant function to help us better understand how the derivatives of the inner and outer functions interact and lead to the final derivative of the composition itself. This use of a derivative is a classic example of the chain rule in calculus, which is a method for determining the derivative of a composite function. The chain rule is vital as it allows the calculation of derivatives for more complex cases and shows up frequently in various disciplines within mathematics and the applied sciences.

Differentiable Functions

A function is considered differentiable at a point if it has a derivative at that point. For a function to have a derivative, it must be continuous; however, continuity alone doesn't guarantee differentiability. A function that is differentiable is smooth without any sharp corners or cusps in its graph at that point. In the exercise, we're given that the functions f and g are differentiable, which implies that they are also continuous. This property is essential because it means that we can apply derivative rules, like the chain rule, without concerns about breaking points or undefined behavior at certain points.

The differentiability of a function provides the assurance needed to predict and understand the function's behavior infinitesimally close to any point within its domain. This insight is powerful, especially when modeling real-world scenarios where predicting change is important, like in physics for motion or in economics for cost analysis.

Composite Functions

A composite function is formed when the output of one function becomes the input of another function. In notation, if we have two functions f and g, then their composite function is written as f(g(x)). The chain rule is the go-to tool when it comes to taking the derivative of such functions. It tells us how to differentiate a composite function systematically by taking the derivative of the outer function evaluated at the inner function and multiplying it by the derivative of the inner function.

The challenge presented in the exercise is to use the chain rule conceptually, represented symbolically rather than through standard rote application. Understanding how to navigate composite functions and their derivatives is paramount as it extends our capability to handle not just simple, but also more dynamically varying systems where one change affects another in a cascading or layered fashion. This ability to dissect complex transformations and handle them one layer at a time is what makes calculus such a powerful tool in scientific and mathematical analysis.