Problem 6
Question
In Exercises \(1-12,\) find \(d y / d x\) $$ y=(\sin x+\cos x) \sec x $$
Step-by-Step Solution
Verified Answer
The derivative \( \frac{dy}{dx} = 1 + \cos x \tan x \).
1Step 1: Apply Product Rule
The given function is a product of two functions: \ \( u(x) = \ ( \sin x + \cos x ) \) and \( v(x) = \sec x \). \ According to the product rule, \( \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \). First, find the derivative of each component.
2Step 2: Differentiate the First Function
Differentiate \( u(x) = \ (\sin x + \cos x) \): \ \( u'(x) = \cos x - \sin x \) using known derivatives \( \frac{d}{dx}(\sin x) = \cos x \) and \( \frac{d}{dx}(\cos x) = -\sin x \).
3Step 3: Differentiate the Second Function
Differentiate \( v(x) = \sec x \): \ \( v'(x) = \sec x \cdot \tan x \) using the derivative of secant: \( \frac{d}{dx}(\sec x) = \sec x \cdot \tan x \).
4Step 4: Apply Product Rule Formula
Now use the product rule formula: \ \[ \frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x) \] \ Substitute the derivatives and original functions: \ \[ \frac{dy}{dx} = (\cos x - \sin x) \cdot \sec x + (\sin x + \cos x) \cdot \sec x \cdot \tan x \]
5Step 5: Simplify the Expression
Simplify the expression by distributing and factoring where possible: \ \[ \frac{dy}{dx} = \cos x \sec x - \sin x \sec x + \sin x \sec x \tan x + \cos x \sec x \tan x \] \ Combine like terms: \ \[ \frac{dy}{dx} = 1 + \cos x \tan x \] simplifying the secant and tangent terms.
Key Concepts
Product RuleTrigonometric DerivativesCalculus Problem Solving
Product Rule
The product rule is a fundamental concept in calculus that allows us to find the derivative of the product of two functions. It is especially handy when you encounter a function like the one in our exercise, which is expressed as the product of two simpler functions.
The rule states that if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product is given by:\[ \frac{d}{dx}(u \, v) = u' \, v + u \, v' \]You basically take the derivative of \( u \) and keep \( v \) as it is, then add the reverse: the original \( u \) and the derivative of \( v \).
To apply it effectively:
The rule states that if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product is given by:\[ \frac{d}{dx}(u \, v) = u' \, v + u \, v' \]You basically take the derivative of \( u \) and keep \( v \) as it is, then add the reverse: the original \( u \) and the derivative of \( v \).
To apply it effectively:
- Identify each function you're multiplying together.
- Differentiate each function.
- Substitute the derivatives into the product rule formula.
- Simplify the expression where possible.
Trigonometric Derivatives
Trigonometric derivatives are the building blocks for finding the rates of change of trigonometric functions, essential for solving calculus problems involving trigonometry.
When dealing with functions like \( \sin x \) and \( \cos x \), or more complex functions like \( \sec x \), it helps to know their derivatives by heart. This knowledge can simplify the differentiation process.
- The derivative of \( \sin x \) is \( \cos x \).- The derivative of \( \cos x \) is \(-\sin x \).- The derivative of \( \sec x \) is \( \sec x \tan x \).If you can remember these, your work in calculus will be much easier. They are often used in conjunction with rules like the product rule or the chain rule to find solutions to more challenging problems.
When dealing with functions like \( \sin x \) and \( \cos x \), or more complex functions like \( \sec x \), it helps to know their derivatives by heart. This knowledge can simplify the differentiation process.
- The derivative of \( \sin x \) is \( \cos x \).- The derivative of \( \cos x \) is \(-\sin x \).- The derivative of \( \sec x \) is \( \sec x \tan x \).If you can remember these, your work in calculus will be much easier. They are often used in conjunction with rules like the product rule or the chain rule to find solutions to more challenging problems.
Calculus Problem Solving
Solving problems in calculus requires a systematic approach to make these complex problems simpler and manageable.
Here's a suggested approach for tackling differentiation problems:
Here's a suggested approach for tackling differentiation problems:
- Understand the problem at hand by identifying the type of functions you are dealing with. Look for patterns and decide which derivative rules apply.
- Use derivative rules such as the product rule, quotient rule, or chain rule to separate the problem into smaller, workable parts.
- Differentiate each part according to its rule. Focus on finding the individual derivatives first before attempting to combine them.
- Combine the derivatives using the rules and simplify whenever possible. Simplification often involves combining like terms and reducing expressions.
Other exercises in this chapter
Problem 6
In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\). $$ y=\sin u, \quad u=x-\cos x $$
View solution Problem 6
In Exercises 1–12, find the first and second derivatives. $$ y=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\frac{x}{4} $$
View solution Problem 6
Exercises \(1-6\) give the positions \(s=f(t)\) of a body moving on a coordinate line, with \(s\) in meters and \(t\) in seconds. a. Find the body's displacemen
View solution Problem 7
Distance Let \(x\) and \(y\) be differentiable functions of \(t\) and let \(s=\sqrt{x^{2}+y^{2}}\) be the distance between the points \((x, 0)\) and \((0, y)\)
View solution