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Chapter 5: Problem 137

State and prove the Chain Rule.

Short Answer

Expert verified
The Chain Rule states that for two functions \(u(x)\) and \(v(u)\), the derivative of the composite function \(v(u(x))\) is given by: \[ (v \circ u)' (x) = v'(u(x)) \cdot u'(x) \] To prove this, we use the definition of a derivative and manipulate the limit expression by multiplying and dividing by \(u(x+h) - u(x)\). This leads to the product of the derivatives of the inner and outer functions, which confirms the Chain Rule: \[ y' = v'(u(x)) \cdot u'(x) \]

Step by step solution

01

Statement of the Chain Rule

The Chain Rule states that if we have two functions say \(u(x)\) and \(v(u)\), then the derivative of the composite function \( v(u(x))\) is given by: \[ (v \circ u)' (x) = v'(u(x)) \cdot u'(x) \] This rule allows us to differentiate complex functions by differentiating the inner and outer functions separately.
02

Proof of the Chain Rule

To prove the chain rule, we will use the definition of a derivative, which states that the derivative of a function \(f(x)\) at a point \(a\) is: \[ f'(a)=\lim_{{h \to 0}} \frac{f(a + h) - f(a)}{h} \] Now, let \(y = v(u(x))\), then the derivative of \(y\) with respect to \(x\) is: \[ y'=\lim _{{h \to 0}} \frac{y(x + h) - y(x)}{h} \] This can be written as: \[ y'=\lim_{{h \to 0}} \frac{v(u(x + h)) - v(u(x))}{h} \] We have a problem here because of \(h\) in the denominator which we can overcome by multiplying and dividing by \(u(x+h) - u(x)\). This gives us: \[ y'=\lim_{{h \to 0}} \frac{v(u(x + h)) - v(u(x))}{u(x+h) - u(x)} \times \frac{u(x+h) - u(x)}{h} \] The first part of the RHS is just the definition of a derivative of \(v(u)\) at \(u(x)\) and the second part of the RHS is the derivative of \(u(x)\). Therefore, we get: \[ y' = v'(u(x)) \cdot u'(x) \] Hence, the Chain Rule is proven.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of a Composite Function
When it comes to calculus, complexity often arises when functions are composed of other functions. The derivative of a composite function is where the Chain Rule plays a crucial role. In simple terms, if you have a function nested inside another, say you're tracking how your garden grows over time and this growth depends on the temperature, which changes throughout the day. You now have a composite function: the growth rate with respect to the changing temperature, and the temperature with respect to time.
Proof of the Chain Rule
Understanding the proof of the Chain Rule can be like watching a magic trick unfold鈥攃omplex at first, but clear once broken down. As seen in the step-by-step solution, the key is crafting a bridge between the inner and outer functions using the definition of a derivative. Imagine you're trying to understand how changing the gear ratio on your bike affects your speed. This change is gradual, and approaching zero, we measure the influence over an infinitesimally small interval. By multiplying and dividing by the natural change in the inner function, we cleverly construct a derivative sandwich that reveals the intimate link between the composite parts. This proof elegantly shows that the rate of change of the composite function is indeed the product of the derivatives of its constituent parts.
Differentiation
Differentiation is the fundamental tool in calculus, enabling us to calculate the rate at which something changes. Formally, it measures how a function's output alters as its input changes by a slight amount. In real-world terms, differentiation helps you understand how quickly your car accelerates when you press the gas pedal, or how rapidly the temperature drops when the sun sets. This concept is what allows for the prediction of behavior in physics, engineering, economics, and innumerable other fields. The Chain Rule, in particular, extends this utility to more complex, layered situations.
Limits in Calculus
At the heart of calculus, and indeed the proof of the Chain Rule, lies the concept of limits. A limit captures the behavior of a function as its inputs inch closer and closer to a certain value. This underpins the entire idea of a derivative, which uses limits to describe instantaneous rates of change. To picture this, think of approaching a traffic light; as you get closer, your decision to stop or go is not based on where you were a block away, but on the limit of your distance to the light as it approaches zero. In calculus, mastering the analysis of limits is foundational for understanding and applying differentiation, and hence, for tackling the full breadth of problems that calculus addresses.

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Most popular questions from this chapter

State and prove the Implicit Function Theorem.

(a) Define "homogeneous of degree \(\mathrm{n}^{\prime \prime}\) and "positively homogeneous of degree \(\mathrm{n}^{\prime \prime}\) for a function of two variables \(\mathrm{F}(\mathrm{x}, \mathrm{y})\) and give a geometric interpretation of homogeneity (b) Determine whether the following functions are homogeneous: (i) \(\mathrm{x}^{2} \mathrm{y} \log (\mathrm{y} / \mathrm{x})\); (ii) \(x^{1 / 3}+x y^{-2 / 3}\); (iii) \(\left[\left(x^{2}-y^{2}\right) /\left(x^{2}+y^{2}\right)\right]\) (iv) \(\mathrm{Ar}^{n}\) where \(\mathrm{A}\) is any constant and \(\mathrm{r}=\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{1 / 2}\)

(a) Let \(\mathrm{f}: \mathrm{R}^{2} \rightarrow \mathrm{R}\) be defined by \(f(x, y)=2 x y\left\\{\left(x^{2}-y^{2}\right) /\left(x^{2}+y^{2}\right)\right\\}, x^{2}+y^{2} \neq 0\) and \(=0, \quad \mathrm{x}=\mathrm{y}=0\). Show that \(\left(\partial^{2} \mathbf{f} / \partial \mathrm{x} \partial \mathrm{y}\right) \neq\left(\partial^{2} \mathrm{f} / \partial \mathrm{x} \partial \mathrm{y}\right)\) and explain why. (b) Does there exist a function \(\mathrm{f}\) with continuous second partial derivatives (i.e., an element of \(\mathrm{C}^{2}\) ) such that of \(/ \partial \mathrm{x}=\mathrm{x}^{2}\) and \partialf \(/ \partial \mathrm{y}=\mathrm{xy}\) ?

Prove Taylor's Theorem for \(\mathrm{f} \in \mathrm{C}^{\mathrm{T}}(\mathrm{E})\) where \(\mathrm{E} \subseteq \mathrm{R}^{\mathrm{n}}\) is an open convex set.

State and prove the Cauchy Mean Value Theorem.
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