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Chapter 16: Problem 32

Find the derivatives of the given functions. \(z(x)=\ln |\csc x+\cot x|\)

Short Answer

Expert verified
The short answer for the derivative of the given function is as follows: \[\frac{dz}{dx}(x) = -\frac{\cos x + \sin x \cos x + \cos^2 x}{\sin^2 x \left|\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right|}\]

Step by step solution

01

Rewrite the function in terms of basic trigonometric functions

First, let's rewrite the given function in terms of sine and cosine functions: \[z(x) = \ln\left|\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right|\]
02

Apply chain rule to find the derivative

Now, let's apply the chain rule to find the derivative of \(z(x)\): Recall that \(\frac{d}{dx} (\ln(g(x)) = \frac{g'(x)}{g(x)}\). Let \(g(x) = \left|\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right|\). To find \(g'(x)\), we need to differentiate \(\frac{1}{\sin x} + \frac{\cos x}{\sin x}\) with respect to \(x\). \(\frac{d}{dx}\left(\frac{1}{\sin x}\right) = -\frac{\cos x}{\sin^2 x}\) (by differentiating the reciprocal of \(\sin x\)) \(\frac{d}{dx}\left(\frac{\cos x}{\sin x}\right) = \frac{-\sin x \cos x - \cos^2 x}{\sin^2 x}\) (by applying the quotient rule) Now, let's add the two derivatives: \[g'(x) = -\frac{\cos x}{\sin^2 x} + \frac{-\sin x \cos x - \cos^2 x}{\sin^2 x}\] Combining the two fractions: \[g'(x) = -\frac{\cos x + \sin x \cos x + \cos^2 x}{\sin^2 x}\] Now, find the derivative of \(z(x)\): \[\frac{dz}{dx}(x) = \frac{-\frac{\cos x + \sin x \cos x + \cos^2 x}{\sin^2 x}}{\left|\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right|}\]
03

Simplify the result

Finally, let's simplify the result: \[\frac{dz}{dx}(x) = -\frac{\cos x + \sin x \cos x + \cos^2 x}{\sin^2 x \left|\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right|}\] This derivative represents the derivative of the given function \(z(x)=\ln |\csc x+\cot x|\).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental tool in calculus for finding the derivative of a composition of functions. It's essential when dealing with functions like this where one function is nested inside another. In simple terms, if you have a function \( h(x) = f(g(x)) \), the chain rule states that the derivative \( h'(x) \) is the derivative of \( f \) with respect to \( g \), times the derivative of \( g \) with respect to \( x \). This is expressed as:
  • \( h'(x) = f'(g(x)) \times g'(x) \)
For our function \( z(x) = \ln|\csc x + \cot x| \), we identified \( g(x) = |\csc x + \cot x| \). Using the chain rule, we first differentiate \( \ln(g(x)) \), which gives \( \frac{g'(x)}{g(x)} \). Applying the chain rule allows us to accurately find the derivative considering the complexity of the composite functions involved.
Trigonometric Functions
Trigonometric functions are vital in calculus and include functions like sine, cosine, tangent, cosecant, secant, and cotangent. Understanding these is crucial, as they often appear in various forms in calculus problems. In trigonometric calculus, certain derivatives are:
  • Sine and Cosine: The derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \(-\sin x \).
  • Secant, Cosecant, and Cotangent: These functions have their derivatives based on the derivatives of sine and cosine. The derivative of \( \csc x \) is \(-\csc x \cot x \), and \( \cot x \) is \(-\csc^2 x \).
When working with our exercise, converting \( \csc x + \cot x \) into their basic sine and cosine forms simplifies the differentiation process using known derivatives.
This understanding is critical when using the quotient rule or chain rule as it helps to break down complex expressions.
Quotient Rule
The quotient rule is another fundamental rule in calculus used to find the derivative of a quotient of two functions. When you have a function \( \frac{u(x)}{v(x)} \), the quotient rule provides a systematic way to differentiate it. The rule states:
  • \( \frac{d}{dx}\left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \)
This rule is especially useful when functions are presented as fractions.
In our problem, we used the quotient rule to differentiate \( \frac{1}{\sin x} + \frac{\cos x}{\sin x} \). By individually deriving each term and then applying the rule, it ensures we account for the division in the calculation. Combining both the derivatives simplifies the process of finding \( g'(x) \) before applying the chain rule.

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