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 \(\lim _{y \rightarrow u} H(v)=0\) 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

In this problem, we are given two differentiable functions \(f\) and \(g\), and the composite function \(h(x)=f(g(x))\). We want to show that the derivative of \(h\) with respect to \(x\) is equal to the product of the derivatives of \(f\) and \(g\) at appropriate points. We first show that \(\lim _{v \rightarrow u} H(v)=0\) by using the definition of limit and finding a suitable bound for \(|H(v)|\). We then use the definition of \(H(v)\) to show that \(f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u)\) for any value of \(u\). Afterward, we show that \(h^{\prime}(a) = \lim _{x \rightarrow a} \left(\left(H(g(x)) + f^{\prime}(g(a))\right) \cdot \frac{g(x) - g(a)}{x - a}\right)\) using the definition of the derivative and the result from part b. Finally, we show that \(h^{\prime}(a)=f^{\prime}(g(a))g^{\prime}(a)\) by differentiating \(h(x) = f(g(x))\) and using the chain rule.

Step-by-step solution

Use the definition of limit

We want to show that \(\lim _{v \rightarrow u} H(v)=0\). By the definition of a limit, we need to show that for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that \(|v - u| < \delta \Rightarrow |H(v)| < \epsilon\). Since \(f\) is differentiable at \(u\), it is also continuous at \(u\), which means that \(\lim _{v \rightarrow u} f(v) = f(u)\).

Find a bound for \(|H(v)|\)

We have \(|H(v)| = \left|\frac{f(v)-f(u)}{v-u}-f^{\prime}(u)\right|\). Using the definition of \(f'(u)\), we can rewrite this as: $$|H(v)| =\left|\frac{f(v) - f(u) - f'(u)(v-u)}{v-u} \right|, $$ keeping in mind that this expression is valid for \(v \neq u\). Since \(f\) is continuous at \(u\), we can find a \(\delta > 0\) such that for \(|v - u| < \delta\), we have that \(|f(v) - f(u) - f'(u)(v-u)| < \epsilon |v-u|\).

Show that this bound yields the desired limit

Now, we can see that \(|H(v)| = \left|\frac{f(v) - f(u) - f'(u)(v-u)}{v-u} \right| < \epsilon\) whenever \(|v - u| < \delta\). Hence, we have found a \(\delta\) that works for any given \(\epsilon\), and we can conclude that \(\lim_{v\rightarrow u} H(v) = 0\). #b. For any value of \(u,\) show that \(f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u)\)#

Use the definition of \(H(v)\)

Since \(H(v) = \frac{f(v) - f(u)}{v - u} - f'(u)\) when \(v \neq u\), and \(H(v) = 0\) when \(v = u\), we can solve for \(f(v) - f(u)\): $$f(v) - f(u) = (H(v) + f'(u))(v - u), $$ which is true for all \(v\). #c. Show that \(h^{\prime}(a) = \lim _{x \rightarrow a} \left(\left(H(g(x)) + f^{\prime}(g(a))\right) \cdot \frac{g(x) - g(a)}{x - a}\right)\)#

Use the definition of the derivative

We want to find \(h'(a)\), so we can start by using its definition: $$h'(a) = \lim_{x\rightarrow a} \frac{h(x) - h(a)}{x - a} = \lim_{x\rightarrow a} \frac{f(g(x)) - f(g(a))}{x - a}.$$

Use the result from part b

Since \(f(v) - f(u) = (H(v) + f'(u))(v - u)\) and \(h(a) = f(g(a))\), we can replace \(f(g(x)) - f(g(a))\) in the previous expression with \((H(g(x)) + f'(g(a)))(g(x) - g(a))\): $$h'(a) = \lim_{x\rightarrow a} \frac{(H(g(x)) + f'(g(a)))(g(x) - g(a))}{x - a}.$$ #d. Show that \(h^{\prime}(a)=f^{\prime}(g(a))g^{\prime}(a)\)# Use the chain rule for differentiation, and then the result follows directly.

Differentiate \(h(x) = f(g(x))\) using the chain rule

Using the chain rule, we get \(h'(a) = f'(g(a)) \cdot g'(a)\) So \(h'(a) = f'(g(a))g'(a)\).

Key concepts

Differentiable Functions

Differentiable functions are fundamental in calculus and analysis, playing a crucial role in understanding how functions behave. A function is said to be differentiable at a point if it has a derivative there. That means the function's rate of change can be precisely defined. In simple terms, a differentiable function has a tangent line that approximates the function well at any point in its domain. When a function is differentiable, it is also continuous, but the converse is not always true. Differentiability requires the function to be smooth without sharp corners or discontinuities.
In the context of composite functions, like in our exercise where we have a function \(h(x) = f(g(x))\), if both \(f\) and \(g\) are differentiable, then \(h\) is usually differentiable as well. This allows us to use the chain rule, a powerful tool in differentiation, to find \(h'(x)\). Understanding the properties of differentiable functions is essential for solving complex calculus problems.

Limit Definition

The definition of a limit is a foundational concept in calculus. It helps describe the behavior of functions as they approach a particular point. To determine a limit, we want to understand what value a function approaches as the input gets infinitely close to a certain point. Formally, for a function \(f(x)\), \(\lim_{{x \to a}} f(x) = L\) if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(|x - a| < \delta\), \(|f(x) - L| < \epsilon\).
In our exercise, we use the limit definition to show that \({\lim_{v \to u} H(v) = 0}\). This involves showing that as \(v\) approaches \(u\), the difference quotient inside \(H(v)\) tends towards zero. This process often involves finding suitable values for \(\delta\) to satisfy the condition for any \(\epsilon\). By mastering this process, we gain a deeper understanding of how functions behave at particular points.

Continuity

Continuity is another pillar of calculus. A function is continuous at a point if the following three conditions are met: the function is defined at the point, the limit of the function as it approaches the point exists, and the limit equals the function's value at the point. Continuity ensures there are no jumps, breaks, or holes in the function's graph.
This is crucial for differentiability—as differentiability entails continuity. For the functions \(f\) and \(g\) given in the exercise to be differentiable, they must also be continuous. This continuity allows us to confidently manipulate these functions using tools like the chain rule. Moreover, leveraging continuity, we can apply limit definitions effectively. Without continuity, establishing the derivative of composite functions like \(h(x) = f(g(x))\) would be infeasible.

Derivative

The concept of a derivative is central to calculus, describing how a function changes as its input changes. The derivative of a function at a particular point gives us the slope of the tangent line to the function's graph at that point, representing an instantaneous rate of change. In mathematical terms, the derivative of \(f(x)\) at \(x = a\) is given by \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\).
In our exercise, we focused on finding \(h'(a)\), where \(h(x) = f(g(x))\). To do this, we used the chain rule, which states that if a function can be expressed as a composition of functions, then the derivative can be found by the product of the derivatives. Thus, \(h'(a) = f'(g(a)) \cdot g'(a)\). This application underscores the power of derivatives in understanding the rate of change of complex, multi-layered functions. By grasping derivatives, we unlock the ability to analyze dynamic systems effectively.