   Chapter 11, Problem 51RE 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. x y − y 2 = x 2 − 3 ; ( − 1 , 1 )

To determine

To calculate: The equation of the tangent to the graph of the equation xyy2=x23 at the point (1,1).

Explanation

Given Information:

The provided equation is xyy2=x23 and the point is (1,1).

Formula used:

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

Product rule of derivative of differentiable functions, f(x) and g(x) is

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x).

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

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, xyy2=x23

Find the slope of tangent line to the equation xyy2=x23 by determining the derivative.

Then, take ddx of both sides of the equation,

ddx(xyy2)=ddx(x23)ddx(xy)ddx(y2)=ddx(x2)ddx(3)

Apply the product rule of derivative and the chain rule,

(dxdxy+xdydx)2y21dydx=ddx(x2)0(dxdxy+xdydx)2ydydx=ddx(x2)0

Now, take the derivative of (x2) which is equal to (2x) and simplify the derivative,

(1y+xdydx)2ydydx=2x0y+xdy

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