Question

How can you use the derivatives \(\frac{d}{d x}(\sin x)=\cos x\) \(\frac{d}{d x}(\tan x)=\sec ^{2} x,\) and \(\frac{d}{d x}(\sec x)=\sec x \tan x\) to remember the derivatives of \(\cos x, \cot x,\) and \(\csc x ?\)

Short answer

Answer: The derivatives of \(\cos x, \cot x\), and \(\csc x\) are: - \(\frac{d}{d x}(\cos x) = -\cos x\) - \(\frac{d}{d x}(\cot x) = -\csc^2 x \sec^2 x + \sec^2 x\) - \(\frac{d}{d x}(\csc x) = -\cot x \cdot \csc x\)

Step-by-step solution

Derivative of \(\cos x\)

To find the derivative of \(\cos x\), note that it's just the negative derivative of \(\sin x\). That is, \(\frac{d}{d x}(\cos x) = -\frac{d}{d x}(\sin x)\). Using the given derivative \(\frac{d}{d x}(\sin x) = \cos x\), we have \(\frac{d}{d x}(\cos x) = -\cos x\).

Derivative of \(\cot x\)

Recall that \(\cot x = \frac{1}{\tan x}\), and we're given that \(\frac{d}{d x}(\tan x) = \sec^2 x\). We can use the chain rule to find the derivative of \(\cot x\). The chain rule states that if \(u = u(v)\) and \(v=v(x)\), then \(\frac{d}{d x}(u(v(x))) = \frac{d u}{d v} \cdot \frac{dv}{dx}\). Using the chain rule for \(\cot x\), let \(u(v) = \frac{1}{v}\) and \(v(x) = \tan x\). Then, we have \(\frac{d u}{d v} = -\frac{1}{v^2}\) and \(\frac{d v}{d x} = \sec^2 x\). So, \(\frac{d}{d x}(\cot x) = \frac{d}{d x}\left(\frac{1}{\tan x}\right) = -\frac{1}{\tan^2 x} \cdot \sec^2 x\). Recall that \(\cot^2 x + 1 = \csc^2 x\), so \(\cot^2 x = \csc^2 x - 1\). This means that: \(\frac{d}{d x}(\cot x) = (-1) (\csc^2 x - 1) \cdot \sec^2 x = -\csc^2 x \sec^2 x + \sec^2 x\).

Derivative of \(\csc x\)

Recall that \(\csc x = \frac{1}{\sin x}\), and we're given that \(\frac{d}{d x}(\sin x) = \cos x\). We can again use the chain rule to find the derivative of \(\csc x\). Using the chain rule for \(\csc x\), let \(u(v) = \frac{1}{v}\) and \(v(x) = \sin x\). Then, we have \(\frac{d u}{d v} = -\frac{1}{v^2}\) and \(\frac{d v}{d x} = \cos x\). So, \(\frac{d}{d x}(\csc x) = \frac{d}{d x}\left(\frac{1}{\sin x}\right) = -\frac{1}{\sin^2 x} \cdot \cos x\). Recall that \(\cot x = \frac{\cos x}{\sin x}\) and \(\csc x = \frac{1}{\sin x}\). This means that: \(\frac{d}{d x}(\csc x) = -\cot x \cdot \csc x\). To summarize, we found the derivatives of \(\cos x, \cot x,\) and \(\csc x\) as follows: - \(\frac{d}{d x}(\cos x) = -\cos x\) - \(\frac{d}{d x}(\cot x) = -\csc^2 x \sec^2 x + \sec^2 x\) - \(\frac{d}{d x}(\csc x) = -\cot x \cdot \csc x\) Using the given derivatives as reference, the student can remember these derivatives by noting their relationships with the trigonometric functions and their derivatives.

Key concepts

Chain Rule in Calculus

The chain rule is a powerful tool in calculus for differentiating composite functions. When a function is a combination of two or more functions, the chain rule comes into play. It allows calculation of the derivative of a composite function by breaking it down into its constituent functions.

For instance, if we have a function in the form of \( u(g(x)) \), then the derivative of this function with respect to \( x \) is given by the product of the derivative of \( u \) with respect to \( g(x) \) and the derivative of \( g(x) \) with respect to \( x \). In mathematical terms, \( \frac{d}{dx}u(g(x)) = \frac{du}{dg} \cdot \frac{dg}{dx} \).

This technique is crucial in finding the derivatives of trigonometric functions that are given in terms of other functions, such as \( \cot x = \frac{1}{\tan x} \) and \( \csc x = \frac{1}{\sin x} \). By carefully applying the chain rule, we simplify the process of finding derivatives for complex functions like these.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables where both sides of the equation are defined. These identities simplify the process of taking derivatives and solving trigonometric equations.

Some important identities that appear frequently in calculus are the Pythagorean identities such as \( \sin^2x + \cos^2x = 1 \) and \( 1 + \cot^2x = \csc^2x \), reciprocal identities like \( \csc x = \frac{1}{\sin x} \), and quotient identities such as \( \tan x = \frac{\sin x}{\cos x} \).

By using these identities, we can transform complex trigonometric expressions into simpler forms that are easier to differentiate. For example, rewriting \( \cot x \) as \( \frac{1}{\tan x} \) or transforming \( \cot^2 x + 1 \) into \( \csc^2 x \) are strategies that leverage trigonometric identities to facilitate differentiation.

Differentiation Techniques

Differentiation is the process of finding the derivative of a function, which represents the rate at which a function is changing at any given point. There are several techniques to differentiate functions, especially when dealing with more complex functions.

Some common differentiation techniques include the power rule, the product rule, the quotient rule, and the aforementioned chain rule. In the scenario of trigonometric functions, we often use these techniques in combination with trigonometric identities.

For example, to find the derivative of \( \cot x \), we reshape it using the quotient identity, and then apply the quotient rule. Similarly, for \( \csc x = \frac{1}{\sin x} \), we use the reciprocal identity and apply the chain rule. Being seasoned in these techniques allows students to tackle a wide variety of functions and find their derivatives systematically and efficiently.