   Chapter 3.5, Problem 31E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use implicit differentiation to find an equation of the tangent line to the curve at the given point.31. 2(x2 + y2)2 = 25(x2 – y2), (3, 1), (lemniscate) To determine

To find: The equation of the tangent line to the given equation at the point.

Explanation

Given:

The curve is 2(x2+y2)2=25(x2y2).

The point is (3,1).

Derivative rule: Chain rule

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydududx.

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (1)

Where, m is the slope of the tangent line at (x1,y1) and m=dydx|(x,y)=(x1,y1).

Calculation:

Consider the given equation 2(x2+y2)2=25(x2y2).

Differentiate the given equation implicitly with respect to x.

ddx(2(x2+y2)2)=ddx(25(x2y2))2ddx(x2+y2)2=25ddx(x2y2)

Let u=x2+y2 and  k=x2y2.

Then, 2ddx(u2)=25ddx(v).

Apply the chain rule and simplify the terms.

2[ddu(u2)dudx]=25[ddv(v)dvdx]2[2ududx]=25[1dvdx]4ududx=25dvdx

Substitute u=x2+y2 and  k=x2y2.

4(x2+y2)ddx(x2+y2)=25ddx(x2y2)4(x2+y2)[ddx(x2)+ddx(y2)]=25[ddx(x2)ddx(y2)]4(x2+y2)[2x+ddx(y2)]=25[2xddx(y2)]

Apply the chain rule and simplify the terms.

4(x2+y2)[2x+2ydydx]=25[2x2ydydx]4(x2+y2)[x+ydydx]=25[xydydx]4x(x2+y2)+4y(x2+y2)dydx=25x25ydydx

Combine the terms dydx to one side of the equation

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