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Chapter 3: Problem 34

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

Short Answer

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Question: Find the derivative of the function $$y=\sec x\tan x$$. Answer: $$y'=\sec x(\sec^2 x+\tan x\tan x)$$

Step by step solution

01

Identify the Functions u(x) and v(x)

In this problem, we need to identify the two functions whose product we must differentiate. We are given $$y=\sec x\tan x$$, so we can identify $$u(x) = \sec x$$ and $$v(x) = \tan x$$.
02

Find the Derivatives of u(x) and v(x)

Now we need to find the derivatives of the functions we identified in step 1. Recall that: $$\frac{d}{dx} \sec x = \sec x \tan x$$ $$\frac{d}{dx} \tan x = \sec^2 x$$ So, we have $$u'(x)=\sec x\tan x$$ and $$v'(x)=\sec^2 x$$.
03

Apply the Product Rule

Now that we have found the derivatives of u(x) and v(x), we can apply the product rule to find the derivative of the function y, which is given by: $$y'=u(x)v'(x) + v(x)u'(x)$$ Substituting the given functions and their derivatives, we get: $$y'=\sec x(\sec^2 x)+\tan x(\sec x\tan x)$$
04

Simplify the Expression

The resulting expression for y' can be simplified by factoring out the common term of $$\sec x$$: $$y'=\sec x(\sec^2 x+\tan x\tan x)$$ Therefore, the derivative of the given function is: $$y'=\sec x(\sec^2 x+\tan x\tan x)$$

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Most popular questions from this chapter

Vertical tangent lines a. Determine the points at which the curve \(x+y^{3}-y=1\) has a vertical tangent line (see Exercise 52 ). b. Does the curve have any horizontal tangent lines? Explain.

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}\left[\frac{f(x)}{(x+2)}\right]\right|_{x=4}$$

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}\left[\frac{f(x) g(x)}{x}\right]\right|_{x=4}$$

A particle travels clockwise on a circular path of diameter \(R,\) monitored by a sensor on the circle at point \(P ;\) the other endpoint of the diameter on which the sensor lies is \(Q\) (see figure). Let \(\theta\) be the angle between the diameter \(P Q\) and the line from the sensor to the particle. Let \(c\) be the length of the chord from the particle's position to \(Q\) a. Calculate \(d \theta / d c\) b. Evaluate \(\left.\frac{d \theta}{d c}\right|_{c=0}\)

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals. . A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Use implicit differentiation if needed to find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=c x^{2} ; x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants
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