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Chapter 3: Problem 8

Find \(d y / d x\). $$g(x)=\frac{\cos x}{\sin ^{2} x}$$

Short Answer

Expert verified
The derivative is \( g'(x) = -\frac{1\text{sin}^2 x + 2\text{cos}^2 x }{\text{sin}^3 x} \).

Step by step solution

01

Rewrite the function

Rewrite the given function in a simpler form to make differentiation easier: \[ g(x) = rac{ ext{cos } x}{( ext{sin } x)^2} = ext{cos } x imes ( ext{sin } x)^{-2} \] This simplification helps us apply the product rule more effectively.
02

Apply the Product Rule

The product rule states: if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product is: \[ \frac{d}{dx}[u imes v] = u' v + u v' \] Here, let \( u(x) = \text{cos } x \) and \( v(x) = ( ext{sin } x)^{-2} \). Differentiate each function individually.
03

Differentiate \( u(x) = \cos x \)

Find the derivative of \( u(x) = \cos x \):\[ u'(x) = -\sin x \]
04

Differentiate \( v(x) = ( ext{sin } x)^{-2} \) using the chain rule

Apply the chain rule to differentiate \( v(x) = ( ext{sin } x)^{-2} \). The derivative is:\[ v'(x) = -2( ext{sin } x)^{-3} \cdot \cos x \] (Differentiate the power using the chain rule: \( n f(x)^{n-1} f'(x) \)).
05

Combine results using the Product Rule

Substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into the product rule formula:\[ g'(x) = (-\sin x) imes ( ext{sin } x)^{-2} + \cos x \times (-2( ext{sin } x)^{-3} \cdot \cos x) \] This simplifies to: \[ g'(x) = -\frac{1}{\text{sin } x} - \frac{2 \text{cos }^2 x}{\text{sin }^3 x} \]
06

Simplify the expression for the derivative

Combine the terms over a common denominator:\[ g'(x) = -\frac{1\text{sin}^2 x + 2\text{cos}^2 x }{\text{sin}^3 x} \]Simplify further if possible by recognizing trigonometric identities or simplification.

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Key Concepts

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

Product Rule in Differentiation
In calculus, the product rule is a vital tool for differentiating functions that are products of two or more functions. When you have a function like \( g(x) = u(x) \cdot v(x) \), you can find its derivative using the product rule:
  • First, identify the two parts, \( u(x) \) and \( v(x) \).
  • Calculate the individual derivatives, \( u'(x) \) and \( v'(x) \).
  • Then, use the formula: \( \frac{d}{dx}[u \cdot v] = u' v + u v' \).
In the given exercise, this rule is particularly helpful because it simplifies the process of finding the derivative of a complex function. Remember, anytime you see a multiplication between two functions that depend on \( x \), think product rule!
Chain Rule Explained
The chain rule is another fundamental method in calculus that allows you to differentiate composite functions. If you have a function \( h(x) = f(g(x)) \), the chain rule helps you by separating the tasks:
  • First, differentiate the outer function \( f \) concerning its input, \( g(x) \).
  • Then, multiply this result by the derivative of the inner function \( g(x) \).
In mathematical terms, it looks like this: \( h'(x) = f'(g(x)) \cdot g'(x) \). In our problem, we applied the chain rule to the portion of the function \( (\sin x)^{-2} \), treating it as an outer function being raised to a power. Using this approach, differentiation becomes manageable even with intricate compositions of functions.
Understanding Trigonometric Functions in Differentiation
Trigonometric functions such as \( \sin x \) and \( \cos x \) are ubiquitous in mathematical problems and often appear in differentiation exercises. They follow specific rules:
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \cos x \) is \(-\sin x \).
These rules are crucial when dealing with any trigonometric components within a function. In our problem, knowing these derivatives allowed us to efficiently apply both the product and chain rules. Understanding how to differentiate trigonometric functions not only simplifies calculations but also strengthens your grasp of calculus concepts.

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Most popular questions from this chapter

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=1-\frac{1}{u}, \quad u=g(x)=\frac{1}{1-x}, \quad x=-1$$

A highway patrol plane flies 3 km above a level, straight road at a steady \(120 \mathrm{km} / \mathrm{h}\). The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is \(5 \mathrm{km},\) the line-of-sight distance is decreasing at the rate of \(160 \mathrm{km} / \mathrm{h}\). Find the car's speed along the highway.

If \(r=\sin (f(t)), f(0)=\pi / 3,\) and \(f^{\prime}(0)=4,\) then what is \(d r / d t\) at \(t=0 ?\)

A weight is attached to a spring and reaches its equilibrium position \((x=0) .\) It is then set in motion resulting in a displacement of $$x=10 \cos t$$ where \(x\) is measured in centimeters and \(t\) is measured in seconds. See the accompanying figure. a. Find the spring's displacement when \(t=0, t=\pi / 3,\) and \(t=3 \pi / 4\) b. Find the spring's velocity when \(t=0, t=\pi / 3,\) and \(t=3 \pi / 4\)

Find \(d y / d t\) $$y=\left(\frac{3 t-4}{5 t+2}\right)^{-5}$$
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