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Chapter 7: Problem 19

Show from first principles that the function cos is differentiable at \(\pi / 2\) and its derivative at \(\pi / 2\) is \(-1\).

Short Answer

Expert verified
Using the first principles of differentiation, we found the derivative for the cos function at \(x = \frac{\pi}{2}\) by evaluating the limit: \[f'(\frac{\pi}{2}) = \lim_{h \to 0} \frac{\cos(\frac{\pi}{2} + h)}{h}\] We then used the sum of angles identity for cosine and simplified the expression. After evaluating the resulting limit, we concluded that the function cos is differentiable at \(\frac{\pi}{2}\) and its derivative at \(\frac{\pi}{2}\) is -1.

Step by step solution

01

Write down the formula for the derivative from first principles

The formula for the derivative of a function f(x) from first principles is given by: \[f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\]
02

Define the function and the point of interest

In this exercise, we are given the function \(f(x) = \cos(x)\) and are asked to find its derivative at \(x = \frac{\pi}{2}\).
03

Substitute the function and point into the derivative formula

We need to substitute the function \(f(x) = \cos(x)\) and the point \(x = \frac{\pi}{2}\) into the derivative formula. Doing this, we get: \[f'(\frac{\pi}{2}) = \lim_{h \to 0} \frac{\cos(\frac{\pi}{2} + h) - \cos(\frac{\pi}{2})}{h}\]
04

Evaluate the function at the point and simplify the expression

We know that \(\cos(\frac{\pi}{2}) = 0\), so we can substitute this value into the expression above: \[f'(\frac{\pi}{2}) = \lim_{h \to 0} \frac{\cos(\frac{\pi}{2} + h) - 0}{h}\] This simplifies to: \[f'(\frac{\pi}{2}) = \lim_{h \to 0} \frac{\cos(\frac{\pi}{2} + h)}{h}\]
05

Use the trigonometric identity for the sum of angles and evaluate the limit

We know that the sum of angles identity for cosine is: \[\cos(A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B\] In our case, \(A = \frac{\pi}{2}\) and \(B = h\). Plugging these values into the identity, we have: \[\frac{\cos(\frac{\pi}{2}+h)}{h} = \frac{\cos(\frac{\pi}{2})\cos(h)-\sin(\frac{\pi}{2})\sin(h)}{h}\] Since \(\cos(\frac{\pi}{2}) = 0\) and \(\sin(\frac{\pi}{2}) = 1\), the expression becomes: \[\frac{\cos(\frac{\pi}{2}+h)}{h} = \frac{-\sin(h)}{h}\] Now, we need to find the limit as \(h\) approaches 0: \[f'(\frac{\pi}{2}) = \lim_{h \to 0} \frac{-\sin(h)}{h}\] This is a standard limit, and it is equal to -1: \[f'(\frac{\pi}{2}) = -1\]
06

Conclusion

Based on the first principles, we've shown that the function cosine is differentiable at \(\frac{\pi}{2}\) and its derivative at \(\frac{\pi}{2}\) is -1.

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Key Concepts

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

Limits in Calculus
The concept of limits is at the heart of calculus. It helps us understand the behavior of functions as they approach a specific point. In practical terms, when we say we are taking the limit of a function as it approaches a point, we're trying to find out what value the function settles on鈥攐r gets arbitrarily close to鈥攁s the input variable gets nearer and nearer to that point.

For example, in the context of differentiability, when we're looking at the derivative of a function at a point, we're actually analyzing the limit of the difference quotient as that point is approached. The standard form of this, as seen in the example with the cosine function, can be written as \[f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\] This formula embodies the core idea of a derivative: the instantaneous rate of change, or the slope, of the tangent line to the function at that specific point.

Understanding how to evaluate such limits, especially when dealing with trigonometric functions鈥攁nd managing cases where the limit isn鈥檛 immediately obvious鈥攊s a key skill in calculus. Often, this involves the application of other mathematical strategies or identities to simplify the limit expression before taking a definitive value.
Trigonometric Identities
Trigonometric identities are equations that involve trigonometric functions and are true for every value of the occurring variables where both sides of the identity are defined. These identities are useful for simplifying complex expressions and solving trigonometric equations.

In the context of the cosine function's differentiability, the trigonometric identity for the sum of angles comes into play: \[\cos(A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B\] In solving our example problem, this identity allowed for the simplification of \(\cos(\frac{\pi}{2} + h)\) into a form that can be managed as h approaches zero. And because the values of \(\cos(\frac{\pi}{2})\) and \(\sin(\frac{\pi}{2})\) are well-known, the limit could be found with relative ease. Recognizing when and how to apply trigonometric identities is a crucial skill for solving problems involving trigonometric functions and their limits.
Derivative From First Principles
The derivative from first principles, also known as the definition of the derivative, is the most fundamental approach to understanding what derivatives represent. It involves the limit of the ratio of the change in the value of a function to the change in a variable. The expression for this principle is: \[f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\]

In our exercise, we used this principle to find the derivative of the cosine function at \(\pi / 2\). It involved a step-by-step approach, combining our understanding of limits, the behavior of the cosine function, and the use of trigonometric identities to simplify our expressions before applying the limit.

By applying the derivative from first principles, not only did we confirm the differentiability of the cosine function at the given point but also determined the value of the derivative. This method reinforces the foundational understanding that the derivative at a point provides the slope of the tangent at that point on the graph of a function, which for the cosine at \(\pi / 2\) turns out to be -1, reflecting the anticipated downward slope at that point on the unit circle.

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Most popular questions from this chapter

Let \(r \in \mathbb{R}\) be positive and consider the function \(f:(0, \infty) \rightarrow \mathbb{R}\) defined by \(f(x)=x^{r}\). Show that the growth rate of \(\ln x\) is less than that of \(f\), while the growth rate of \(\exp x\) is more than that of \(f\) as \(x \rightarrow \infty\).

Let \(\alpha, \beta \in \mathbb{R}\). Suppose \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) are differentiable functions such that $$ f^{\prime}=\alpha f+\beta g, \quad g^{\prime}=\alpha g-\beta f, \quad f(0)=0, \quad \text { and } \quad g(0)=1 $$ Show that \(f(x)=e^{\alpha x} \sin \beta x\) and \(g(x)=e^{\alpha x} \cos \beta x\) for all \(x \in \mathbb{R}\). (Hint: Consider \(h: \mathbb{R} \rightarrow \mathbb{R}\) given by \(h(x):=\left(f(x)-e^{\alpha x} \sin \beta x\right)^{2}+(g(x)-\) \(\left.e^{\alpha x} \cos \beta x\right)^{2} .\) Find \(\left.h^{\prime} .\right)\) (Compare Exercise 4.7.)

Let \(\alpha, \beta \in \mathbb{R}\). Suppose \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) are differentiable functions such that $$ f^{\prime}=\alpha f+\beta g, \quad g^{\prime}=\alpha g+\beta f, \quad f(0)=0, \quad \text { and } \quad g(0)=1 $$ Show that \(f(x)=e^{\alpha x} \sinh \beta x\) and \(g(x)=e^{\alpha x} \cosh \beta x\) for all \(x \in \mathbb{R}\).

Prove the following for all \(x_{1}, x_{2} \in \mathbb{R}\) : $$ \begin{aligned} &2 \sin x_{1} \cos x_{2}=\sin \left(x_{1}+x_{2}\right)+\sin \left(x_{1}-x_{2}\right) \\ &2 \cos x_{1} \cos x_{2}=\cos \left(x_{1}+x_{2}\right)+\cos \left(x_{1}-x_{2}\right) \\ &2 \sin x_{1} \sin x_{2}=\cos \left(x_{1}-x_{2}\right)-\cos \left(x_{1}+x_{2}\right) . \end{aligned} $$ Deduce the following orthogonality relations: For every \(k, j \in \mathbb{Z}\), $$ \int_{-\pi}^{\pi}(\sin k x)(\cos j x) d x=0 $$ and if \(k \neq j\), then $$ \int_{-\pi}^{\pi}(\cos k x)(\cos j x) d x=0=\int_{-\pi}^{\pi}(\sin k x)(\sin j x) d x . $$

Let \(n \in \mathbb{N}\) and consider the function \(f_{n}: \mathbb{R} \rightarrow \mathbb{R}\) defined by $$ f_{n}(x):=\left\\{\begin{array}{ll} x^{n} \sin (1 / x) & \text { if } x \neq 0 \\ 0 & \text { if } x=0 \end{array}\right. $$ Prove the following: (i) If \(n\) is odd and \(k:=(n-1) / 2\), then \(f_{n}^{(k)}\) exists and is continuous on \(\mathbb{R}\), but \(f_{n}^{(k+1)}\) does not exist at 0 . (ii) If \(n\) is even and \(k:=n / 2\), then \(f_{n}^{(k)}\) exists on \(\mathbb{R}\), but it is not continuous at \(0 .\) (Compare Exercise 4.13.)
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