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Chapter 3: Problem 34

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

Short Answer

Expert verified
Question: Given the function \(y = \sec{x}\tan{x}\), find its derivative with respect to x. Answer: The derivative, \(\frac{dy}{dx}\), is \(\sec x \tan^2 x + \sec^3 x\).

Step by step solution

01

Identify the functions and their derivatives

In this problem, we have: $$f(x) = \sec x, \qquad g(x) = \tan x$$ We need to find their derivatives: $$f'(x) = \frac{d(\sec x)}{dx} = \sec x \tan x$$ $$g'(x) = \frac{d(\tan x)}{dx} = \sec^2 x$$
02

Apply the product rule

Now that we have the functions and their derivatives, we'll use the product rule: \((fg)' = f'g + fg'\). Applying the product rule for our functions, we get: $$\frac{d}{dx}(\sec x \tan x) = (\sec x \tan x)(\tan x) + (\sec x)(\sec^2 x)$$
03

Simplify the derivative

Simplify the derivative expression: $$\frac{d}{dx}(\sec x \tan x) = \sec x \tan^2 x + \sec^3 x$$ Finally, we have found the derivative of the given function: $$\frac{d}{dx}(\sec x \tan x) = \sec x \tan^2 x + \sec^3 x$$

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Most popular questions from this chapter

Rule In general, the derivative of a quotient is not the quotient of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f / g\) equals \(f^{\prime} / g^{\prime}\)

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An observer is \(20 \mathrm{m}\) above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is \(20 \mathrm{m}\) horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of \(5 \mathrm{m} / \mathrm{s}\), what is the rate of change of the angle of elevation when the elevator is \(10 \mathrm{m}\) above the ground? When the elevator is \(40 \mathrm{m}\) above the ground?
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