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Magazine Chapter 3: Problem 26
Assuming that \((d / d x)(\sin x)=\cos x\) and \((d / d x)(\cos x)=\) \(-\sin x,\) prove each of the following. (a) \(\frac{d}{d x} \cot x=-\csc ^{2} x\) (b) \(\frac{d}{d x} \csc x=-\csc x \cot x\)
Short Answer
Expert verified
Part (a): \(\frac{d}{d x} \cot x = -\csc ^{2} x\). Part (b): \(\frac{d}{d x} \csc x = -\csc x \cot x\).
Step by step solution
01
Part (a) Step 1: Rewrite \(\cot x\) as a quotient
\(\cot x\) can be expressed as \(\frac{1} {\tan x}\), which itself can be expressed as \(\frac{\cos x} {\sin x}\). So, the task now is to differentiate \(\frac{\cos x} {\sin x}\).
02
Part (a) Step 2: Use quotient rule to differentiate
The quotient rule states that for a function of the form \(\frac{u}{v}\) its derivative is \(\frac{v (du) / dx - u (dv) / dx} {v^2}\). By applying this rule to \(\frac{\cos x} {\sin x}\), we get: \(\frac{\sin x *(d/dx)(\cos x) - \cos x *(d/dx)(\sin x)} {(\sin x)^2}\). We can substitute \(cos x\) for \((d/dx)(\sin x)\) and \(-\sin x\) for \((d/dx)(\cos x)\). This yields \(\frac{-\sin ^2 x - \cos ^2 x} {\sin ^2 x}\).
03
Part (a) Step 3: Simplify
We use the Pythagorean identity \(\sin ^2 x + \cos ^2 x = 1\) to simplify the numerator to \(-1\). Then, the expression becomes \(-\frac{1} {\sin ^2 x}\), which is \(-\csc ^2 x\). This completes the proof for part (a).
04
Part (b) Step 1: Rewrite \(\csc x\) as a reciprocal
\(\csc x\) can be written as \(\frac{1} {\sin x}\). Our task now is to differentiate this expression.
05
Part (b) Step 2: Use chain rule to differentiate
The chain rule states that if \(y = u(v)\), then \(dy/dx = (du/dv)*(dv/dx)\). Here, we consider \(\frac{1} {\sin x}\) as \(u(v) = \frac{1} {v}\) where \(v = \sin x\). Differentiating \(u(v)\) with respect to \(v\) gives \(-\frac{1} {v^2}\), and the derivative of \(\sin x\) is \(\cos x\). By substituting these and multiplying, we have \(-\frac{\cos x} {\sin ^2 x}\).
06
Part (b) Step 3: Simplify
Expressing \(\frac{1} {\sin x}\) as \(\csc x\) and \(\frac{\cos x} {\sin x}\) as \(\cot x\), we simplify the expression to \(-\csc x \cot x\). This completes the proof for part (b).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative of Trigonometric Functions
Trigonometric functions are fundamental elements of calculus, and their derivatives are essential for solving a variety of problems.
When finding the derivatives of trigonometric functions, certain rules and identities are used. For instance, the derivative of \(sin x\) is \(cos x\), and the derivative of \(cos x\) is \( -sin x\). These are standard results that one can utilize when differentiating more complex trigonometric expressions.
Understanding these basic derivatives allows us to tackle more complicated functions, such as \(cot x\) and \(csc x\), by expressing them in terms of \(sin x\) and \(cos x\), and then applying differentiation rules.
When finding the derivatives of trigonometric functions, certain rules and identities are used. For instance, the derivative of \(sin x\) is \(cos x\), and the derivative of \(cos x\) is \( -sin x\). These are standard results that one can utilize when differentiating more complex trigonometric expressions.
Understanding these basic derivatives allows us to tackle more complicated functions, such as \(cot x\) and \(csc x\), by expressing them in terms of \(sin x\) and \(cos x\), and then applying differentiation rules.
Quotient Rule
The quotient rule is a technique for differentiating functions that are divided by each other, essentially fractions of functions.
Suppose you have a function \(y = \frac{u}{v}\), where both \(u\) and \(v\) are themselves functions of \(x\). According to the quotient rule, the derivative \(y'\) (or \(\frac{dy}{dx}\)) is given by \(\frac{v(du/dx) - u(dv/dx)}{v^2}\).
When applying the quotient rule, it is crucial to carefully keep track of the numerator \(u\) and the denominator \(v\), as well as their respective derivatives \(du/dx\) and \(dv/dx\), to correctly determine the derivative of the quotient.
Suppose you have a function \(y = \frac{u}{v}\), where both \(u\) and \(v\) are themselves functions of \(x\). According to the quotient rule, the derivative \(y'\) (or \(\frac{dy}{dx}\)) is given by \(\frac{v(du/dx) - u(dv/dx)}{v^2}\).
When applying the quotient rule, it is crucial to carefully keep track of the numerator \(u\) and the denominator \(v\), as well as their respective derivatives \(du/dx\) and \(dv/dx\), to correctly determine the derivative of the quotient.
Chain Rule
The chain rule is a powerful tool in differential calculus for finding the derivative of a composite function.
Imagine you have two functions \(u(x)\) and \(v(u)\), and you want to find the derivative of the composite function \(y = v(u(x))\). The chain rule tells us that the derivative \(dy/dx\) is the product of \(dv/du\) and \(du/dx\). In simple terms, you differentiate the outer function \(v\) with respect to \(u\) and multiply this by the derivative of the inner function \(u\) with respect to \(x\).
This rule is especially useful when dealing with functions inside other functions, like \(1/sin x\) which can be seen as \(1/u\) where \(u = sin x\), allowing for systematic differentiation.
Imagine you have two functions \(u(x)\) and \(v(u)\), and you want to find the derivative of the composite function \(y = v(u(x))\). The chain rule tells us that the derivative \(dy/dx\) is the product of \(dv/du\) and \(du/dx\). In simple terms, you differentiate the outer function \(v\) with respect to \(u\) and multiply this by the derivative of the inner function \(u\) with respect to \(x\).
This rule is especially useful when dealing with functions inside other functions, like \(1/sin x\) which can be seen as \(1/u\) where \(u = sin x\), allowing for systematic differentiation.
Pythagorean Identity
One of the most well-known identities in trigonometry is the Pythagorean identity, which states \(sin^2 x + cos^2 x = 1\).
This identity is helpful when simplifying trigonometric expressions during differentiation. For example, when you encounter sums or differences of squares of sine and cosine, you can often replace them with simpler terms to make the calculus operations easier.
Utilizing this identity effectively can transform the differentiation of more complex trigonometric functions into a more manageable task by reducing the function into terms involving only \(sin\) or \(cos\), as seen during the simplification step in the differentiation of \(cot x\) and \(csc x\).
This identity is helpful when simplifying trigonometric expressions during differentiation. For example, when you encounter sums or differences of squares of sine and cosine, you can often replace them with simpler terms to make the calculus operations easier.
Utilizing this identity effectively can transform the differentiation of more complex trigonometric functions into a more manageable task by reducing the function into terms involving only \(sin\) or \(cos\), as seen during the simplification step in the differentiation of \(cot x\) and \(csc x\).
