   Chapter 3.5, Problem 27E ### 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.27. x2 – xy – y2 = 1, (2, 1) (hyperbola)

To determine

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

Explanation

Given:

The equation is x2xyy2=1.

The point is (2,1).

Derivative rules:

(1) Chain rule: If y=f(u) and u=g(x)  are both differentiable function, then

dydx=dydududx.

(2) Product rule: If y=f(u) and u=g(x)  are both differentiable function, then

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

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=x1,y=y1.

Calculation:

Consider the equation x2xyy2=1.

Differentiate the above equation implicitly with respect to x,

ddx(x2xyy2)=ddx(1)ddx(x2)ddx(xy)ddx(y2)=0

Apply the product rule (2),

2x[xddx(y)+yddx(x)]ddx(y2)=02x[xdydx+y(

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