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Find the derivative of the following functions. $$y=\sec x+\csc x$$

Short Answer

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Question: Find the derivative of the function $$y = \sec x + \csc x$$. Answer: The derivative of the given function is $$y' = \sec x \cdot \tan x - \csc x \cdot \cot x$$.

Step by step solution

01

Differentiate each term individually

First, we will differentiate each term in the function. For this purpose, we will use the following derivative rules: $$\frac{d}{dx}(\sec x) = \sec x \cdot \tan x$$ $$\frac{d}{dx}(\csc x) = - \csc x \cdot \cot x$$
02

Apply the sum rule

Now, we need to use the sum rule for differentiation, which states that the derivative of a sum of functions is the sum of the derivatives of the individual functions: $$(f(x) + g(x))' = f'(x) + g'(x)$$
03

Find the derivative of the given function

Following the sum rule, we have: $$\frac{dy}{dx} = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\csc x) = \sec x \cdot \tan x -\csc x \cdot \cot x$$ So, the derivative of the given function is: $$y' = \sec x \cdot \tan x -\csc x \cdot \cot x$$

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Key Concepts

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

Derivative Rules
Differentiation is a fundamental tool in calculus that measures how a function changes as its input changes. The derivative of a function at a point is a measure of the rate at which the function's value changes with respect to changes in the input. Various rules simplify the process of differentiation when dealing with different types of functions, including trigonometric functions.

Common derivative rules include:
  • Power Rule: For a function of the form \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
  • Product Rule: For the product of two functions \( f(x) = u(x)v(x) \), the derivative is \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
  • Quotient Rule: For a ratio of two functions \( f(x) = \frac{u(x)}{v(x)} \), the derivative is \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \).
  • Chain Rule: For a composite function \( f(x) = g(h(x)) \), the derivative is \( f'(x) = g'(h(x)) \times h'(x) \).
When working with trigonometric functions like \( \text{sec}(x) \) and \( \text{csc}(x) \), we need to remember their specific derivative forms. The derivative of the secant function is \( \frac{d}{dx}(\text{sec} x) = \text{sec} x \times \text{tan} x \), and for cosecant, it's \( \frac{d}{dx}(\text{csc} x) = -\text{csc} x \times \text{cot} x \). Knowing these rules allows for systematic calculation of derivatives for complex expressions.
Sum Rule for Differentiation
The sum rule is a principle in calculus that greatly simplifies the differentiation process when we are dealing with sums of functions. The rule states that the derivative of a sum of two or more functions is simply the sum of the derivatives of the individual functions. Mathematically, we express it as:
\[ (f(x) + g(x))' = f'(x) + g'(x) \]
This rule is both powerful and intuitive because it tells us that to find the derivative of a sum, we can independently differentiate each term and then add the results together.

For example, if we have a function like \( y = \text{sec}(x) + \text{csc}(x) \), we differentiate each term to find:\[ \frac{dy}{dx} = \frac{d}{dx}(\text{sec} x) + \frac{d}{dx}(\text{csc} x) \]
Applying this rule makes differentiation of complex expressions straightforward and manageable. It is essential for students to remember this rule as it frequently appears in calculus problems.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where both sides of the equation are defined. These identities are vital in simplifying trigonometric expressions and in solving equations involving trigonometric functions.

There are several categories of these identities:
  • Pythagorean Identities: such as \( \text{sin}^2(x) + \text{cos}^2(x) = 1 \).
  • Reciprocal Identities: which include \( \text{sec}(x) = \frac{1}{\text{cos}(x)} \) and \( \text{csc}(x) = \frac{1}{\text{sin}(x)} \).
  • Angle Sum and Difference Identities: used to express functions of sum or difference of angles in terms of products of functions of single angles, such as \( \text{sin}(x \text{±} y) = \text{sin} x \text{cos} y \text{±} \text{cos} x \text{sin} y \).
  • Double Angle and Half Angle Identities: which relate the functions of double or half angles to the functions of single angles.
Understanding these identities allows for the manipulation and simplification of trigonometric functions which is critical for solving more complex calculus problems. These identities also establish a foundation for integrating trigonometric functions and for solving differential equations involving trigonometric expressions.

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Most popular questions from this chapter

Tangent lines and exponentials Assume \(b\) is given with \(b > 0\) and \(b \neq 1 .\) Find the \(y\) -coordinate of the point on the curve \(y=b^{x}\) at which the tangent line passes through the origin. (Source: The College Mathematics Journal, \(28,\) Mar 1997 )

Tangency question It is easily verified that the graphs of \(y=x^{2}\) and \(y=e^{x}\) have no point of intersection (for \(x>0\) ), while the graphs of \(y=x^{3}\) and \(y=e^{x}\) have two points of intersection. It follows that for some real number \(2 < p < 3,\) the graphs of \(y=x^{p}\) and \(y=e^{x}\) have exactly one point of intersection (for \(x > 0) .\) Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\). b. Find the slope of the curve at the given point. $$x^{2 / 3}+y^{2 / 3}=2 ;(1,1)$$

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u \end{array}\right.$$ a. Show that \(\lim _{y \rightarrow u} H(v)=0\) b. For any value of \(u,\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u)$$ c. Show that $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right)$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\)

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