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$$ \begin{array}{ll}{\text { Find } y^{\prime \prime} \text { if }} \\ {\text { a. } y=\csc x} & {\text { b. } y=\sec x}\end{array} $$

Short Answer

Expert verified
\( y'' = \csc x (\cot^2 x + \csc^2 x) \) for \( y = \csc x \); \( y'' = \sec x (\sec^2 x + \tan^2 x) \) for \( y = \sec x \).

Step by step solution

01

Differentiate the first function

To find the second derivative of \( y = \csc x \), start by finding the first derivative \( y' \). Recall the derivative of \( \csc x \) is \( y' = -\csc x \cot x \).
02

Differentiate the second function

We need to differentiate again to find \( y'' \) for \( y = \csc x \). Apply the product and chain rules to obtain \( y'' = -\csc x (-\csc^2 x) - \csc x \cot^2 x = \csc x (\cot^2 x + \csc^2 x) \).
03

Differentiate the first function for part b

For \( y = \sec x \), differentiate to find \( y' \). The derivative of \( \sec x \) is \( y' = \sec x \tan x \).
04

Differentiate the second function for part b

Differentiate \( y' = \sec x \tan x \) using the product rule: \( y'' = \sec x(\sec^2 x) + \tan x(\sec x \tan x) \). Simplified, this becomes \( y'' = \sec x (\sec^2 x + \tan^2 x) \).

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Key Concepts

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

Cosecant Function
The cosecant function, denoted as \( \csc x \), is one of the six trigonometric functions. It is the reciprocal of the sine function. This means that \( \csc x = \frac{1}{\sin x} \). Because sine varies between -1 and 1, the cosecant function can take any value except for 0, wherever sine is not 0. The graph of the cosecant function looks like a series of curves divided by vertical asymptotes where \( \sin x = 0 \). These asymptotes occur at integer multiples of \( \pi \).

In calculus, finding derivatives involving \( \csc x \) often requires utilizing identities and simplifications. For example, its first derivative is \( y' = -\csc x \cot x \), which can be derived from the quotient rule or by considering it as the product of \(-\sin x\) and \(\cot x \). Understanding the behavior and properties of \( \csc x \) is crucial for solving related problems in calculus.
Secant Function
The secant function, denoted as \( \sec x \), is another essential trigonometric function and is the reciprocal of the cosine function. Thus, \( \sec x = \frac{1}{\cos x} \). Similar to cosecant, it is undefined where its reciprocal (cosine) is zero, which happens at odd multiples of \( \frac{\pi}{2} \).

The secant function's graph features U-shaped curves between these undefined points, and it also extends infinitely high and low as it approaches these points. For calculus applications, recognizing these characteristics helps when differentiating or integrating expressions involving secant. Calculating the first derivative of \( \sec x \) yields \( y' = \sec x \tan x \), highlighting the close relationship between secant and tangent functions.

As we progress to the second derivative, expressions become more complex, often involving both \( \sec^2 x \) and \( \tan^2 x \), requiring careful application of differentiation techniques like the product rule.
Product Rule
The product rule in calculus is a vital technique used to differentiate products of two functions. If you have a function \( y = u(x) \cdot v(x) \), the derivative \( y' \) can be found using the formula:
  • \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)
Essentially, the rule involves differentiating each function separately while multiplying by the other original function, then summing the results. This rule plays a crucial role when differentiating functions like \( y = \sec x \tan x \), where both functions, \( u(x) = \sec x \) and \( v(x) = \tan x \), need individual consideration.

In practice, applying the product rule requires careful attention to detail, especially when combined with other rules like the chain rule. Calculating the second derivative may further complicate things, especially as intermediate derivatives need further differentiation.
Chain Rule
The chain rule allows us to find the derivative of composite functions. A composite function is one that can be described as \( y = f(g(x)) \). To differentiate such a function, we apply the chain rule:
  • \( y' = f'(g(x)) \cdot g'(x) \)
This involves taking the derivative of the outer function \( f \) evaluated at the inner function \( g(x) \), then multiplying it by the derivative of the inner function \( g(x) \).

This rule is particularly useful when dealing with nested trigonometric functions, as seen with \( \csc x \) and \( \sec x \). When differentiating these functions for the second derivative, the chain rule often works hand-in-hand with the product rule to handle more complex expressions. Mastery of the chain rule simplifies handling derivatives of intricate expressions and is crucial for advancing in calculus.

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Most popular questions from this chapter

Use a CAS to perform the following steps in Exercises \(77-84\) . a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P .\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P\) . Then plot the implicit curve and tangent line together on a single graph. $$ y^{3}+\cos x y=x^{2}, \quad P(1,0) $$

In Exercises \(81-86,\) find a parametrization for the curve. the ray (half line) with initial point \((-1,2)\) that passes through the point \((0,0)\)

Quadratic approximations Let \(Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2}\) be a quadratic approximation to \(f(x)\) at \(x=a\) with the properties: i. \(Q(a)=f(a)\) ii. \(Q^{\prime}(a)=f^{\prime}(a)\) iii. \(Q^{\prime \prime}(a)=f^{\prime \prime}(a)\) Determine the coefficients \(b_{0}, b_{1},\) and \(b_{2}\) b. Find the quadratic approximation to \(f(x)=1 /(1-x)\) at \(x=0\) . c. Graph \(f(x)=1 /(1-x)\) and its quadratic approximation at \(x=0 .\) Then zoom in on the two graphs at the point \((0,1)\) . Comment on what you see. d. Find the quadratic approximation to \(g(x)=1 / x\) at \(x=1\) . Graph \(g\) and its quadratic approximation together. Comment on what you see. e. Find the quadratic approximation to \(h(x)=\sqrt{1+x}\) at \(x=0 .\) Graph \(h\) and its quadratic approximation together. Comment on what you see. f. What are the linearizations of \(f, g,\) and \(h\) at the respective points in parts (b), (d), and (e)?

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ 2 x y+\pi \sin y=2 \pi, \quad(1, \pi / 2) $$

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ x^{2}+y^{2}=25, \quad(3,-4) $$

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