   Chapter 8.4, Problem 44E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding an Equation of a Tangent Line In Exercises 43–50, find an equation of the tangent line to the graph of the function at the given point. Function Point y = sec   x ( π 3 ,   2 )

To determine

To calculate: The equation of the tangent line to the graph of the provided function y=secx at given point (π3,2).

Explanation

Given Information:

The provided function is y=secx at given point (π3,2).

Formula used:

Secant differentiation rule:

ddx[secu]=secutanududx

General power rule of differentiation:

ddx[xn]=nxn1

Write trigonometric reciprocal identities.

secθ=1cosθtanθ=sinθcosθ

Write the equation of the tangent line at a given point (x1,y1).

yy1=dydx(x1,y1)(xx1)

Here, dydx(x1,y1) represents slope at point (x1,y1).

Calculation:

Consider the provided function:

y=secx

Differentiate the above function using Secant and general power rules of differentiation.

dydx=ddx[secx]=secxtanx

The derivative of function y=secx

dydx=secxtanx

Substitute, π3 for x in above derivative

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