   Chapter 11, Problem 48RE Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Solutions

Chapter
Section Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

In Exercises 47-52, find the equation of the tangent line to the graph of the given equation at the specified point. y = ( 2 x 2 − 3 ) − 3 ; x = − 1

To determine

To calculate: The equation of the tangent to the graph of the equation y=(2x23)3 at x=1.

Explanation

Given Information:

The provided equation is y=(2x23)3 and x=1.

Formula used:

Slope of tangent to the graph of f(x) at point (x1,y1) is given by dydx|(x1,y1).

The derivative of function f(x)=un using the chain rule is f(x)=ddx(un)=nun1dudx.

Where, u is the function of x.

Constant multiple rule of derivative of function f(x) is f'(cx)=cf'(x) where c is constant.

Equation of line is y=mx+b.

Where, m is the slope and b=y1mx1 when line passes through (x1,y1).

Calculation:

Consider the equation, y=(2x23)3

Find the slope of tangent line to the equation y=(2x23)3 by determining the derivative.

Then, take ddx of both sides of the equation,

ddx(y)=ddx(2x23)3dydx=ddx(2x23)3

Apply the chain rule,

dydx=3(2x23)31ddx(2x23)=3(2x23)4(ddx(2x2)ddx(3))

Now, apply the constant multiple rule,

dydx=3(2x23)4(2ddx(x2)0)=3(2x23)4(2ddx(x2))=6(2x23)4ddx(x2)

Then, take the derivative of (x2) which is equal to (2x),

dydx=6(2x23)42x=12x(2x23)4

So, the slope of the tangent to the graph of equation y=(2x23)3 is 12x(2x23)4

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