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Find \(d y / d x\). $$y=(\sin x+\cos x) \sec x$$

Short Answer

Expert verified
\( \frac{d y}{d x} = \sec^2 x \)

Step by step solution

01

Simplify the expression

Start by rewriting the given expression for easier differentiation: \( y = \sec x \sin x + \sec x \cos x \). Now, since \( \sec x = \frac{1}{\cos x} \), replace \( \sec x \) in both terms. The expression becomes \( y = \frac{\sin x}{\cos x} + 1 \).
02

Differentiate each term separately

Differentiate the expression \( y = \frac{\sin x}{\cos x} + 1 \) with respect to \( x \). The derivative of a constant, like 1, is 0. Use the quotient rule to differentiate \( \frac{\sin x}{\cos x} \), where \( u = \sin x \) and \( v = \cos x \).
03

Apply the quotient rule

The quotient rule formula is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \). For \( u = \sin x \) and \( v = \cos x \), \( \frac{du}{dx} = \cos x \) and \( \frac{dv}{dx} = -\sin x \).
04

Substitute derivatives into the quotient rule

Using the derivatives found previously, substitute into the quotient rule: \( \frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) = \frac{\cos x (\cos x) - \sin x (-\sin x)}{(\cos x)^2} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \).
05

Simplify the derivative expression

Since \( \cos^2 x + \sin^2 x = 1 \) (from the Pythagorean identity), the expression simplifies to \( \frac{1}{\cos^2 x} = \sec^2 x \).
06

Write the final derivative result

Combine the derivative of the constant \(0\) with the result from step 5: \( \frac{d y}{d x} = \sec^2 x \).

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Key Concepts

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

Quotient Rule
The Quotient Rule is a vital tool in calculus when differentiating ratios of two functions. When you have a function like \( \frac{u}{v} \), where \( u \) and \( v \) are both differentiable functions of \( x \), you'll need the Quotient Rule to correctly find its derivative. This rule states:
  • \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)
To apply this, you need to find derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), and then substitute these into the equation. Let’s take an example where \( u = \sin x \) and \( v = \cos x \). Here, \( \frac{du}{dx} = \cos x \) and \( \frac{dv}{dx} = -\sin x \), which we substitute back into the formula to get:
  • \( \frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) = \frac{\cos x (\cos x) - \sin x (-\sin x)}{\cos^2 x} \)
Thus, the specific calculation gives a simplified form using well-known identities, demonstrating how powerful the rule is in breaking down complex functions into simple derivatives.
Trigonometric Identities
Trigonometric identities simplify the process of differentiation, especially for functions involving sine, cosine, and other trigonometric terms. These identities are fundamental relationships that hold between various trigonometric functions.
  • One crucial identity is the Pythagorean identity: \( \cos^2 x + \sin^2 x = 1 \). This identity facilitates simplifying derivative expressions.
  • In solving derivatives like \( \frac{1}{\cos^2 x} \), recognizing that the numerator \( \cos^2 x + \sin^2 x = 1 \) allows us to simplify expressions efficiently.
By applying these identities, difficult problems become manageable. For example, in this exercise, the identity transforms a complicated expression into the simpler \( \sec^2 x \), showing how trigonometric identities underpin and ease differentiation in calculus.
Calculus Differentiation
Differentiation is the process that allows us to find the rate of change of a function with respect to a variable, typically \( x \). In calculus, differentiation involves generating the derivative of a function, which provides insights into how the function behaves.
  • The derivative \( \frac{dy}{dx} \), represents the slope of the tangent line at any point on the curve \( y \).
  • This concept is particularly powerful when applied to functions with trigonometric components, like \( y = (\sin x + \cos x) \sec x \).
  • By systematically applying rules like the product, quotient, and chain rules, even complex expressions yield manageable results, such as the transformation from a layered expression into \( \sec^2 x \).
Understanding these principles allows deeper exploration of the functions' behavior, equipping students with tools to solve a vast array of mathematical and real-life problems.

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Most popular questions from this chapter

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.

A spherical iron ball \(8 \mathrm{cm}\) in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of \(10 \mathrm{cm}^{3} / \mathrm{min},\) how fast is the thickness of the ice decreasing when it is \(2 \mathrm{cm}\) thick? How fast is the outer surface area of ice decreasing?

Find \(d y / d t\) $$y=4 \sin (\sqrt{1+\sqrt{t}})$$

Using the Chain Rule, show that the Power Rule \((d / d x) x^{n}=n x^{n-1}\) holds for the functions \(x^{n}\). $$x^{1 / 4}=\sqrt{\sqrt{x}}$$

In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Würzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about \(7 \mathrm{L} / \mathrm{min} .\) At rest it is likely to be a bit under \(6 \mathrm{L} / \mathrm{min}\). If you are a trained marathon runner running a marathon, your cardiac output can be as high as \(30 \mathrm{L} / \mathrm{min}\). Your cardiac output can be calculated with the formula $$y=\frac{Q}{D},$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{mL} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{mL} / \mathrm{min}\) and \(D=97-56=41 \mathrm{mL} / \mathrm{L}\), \(y=\frac{233 \mathrm{mL} / \mathrm{min}}{41 \mathrm{mL} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}\), fairly close to the \(6 \mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

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