   Chapter 8.4, Problem 47E ### 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
1 views

# 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 = cot 3   x ( π 4 ,   1 )

To determine

To calculate: The equation of the tangent line to the graph of the provided function y=cot3x at given point (π4,1).

Explanation

Given Information:

The provided function is y=cot3x at given point (π4,1).

Formula used:

Cotangent differentiation rule:

ddx[cotu]=csc2ududx

General power rule of differentiation:

ddx[xn]=nxn1

Write trigonometric reciprocal identities.

cscθ=1sinθcotθ=cosθsinθ

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=cot3x

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

dydx=ddx[cot3x]=3cot2csc2x

The derivative of function y=cot3x

dydx=3cot2csc2x

Substitute, π4 for x in above derivative.

dydx(π4,1)==3cot2(π4)csc2(π4)=3(cos2(π4)sin2(π4))(1sin2(π4))=3(cos(π4)cos(π4)sin(π4)sin(π4))(1sin(π4)sin(π

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 