BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.5, Problem 23E
To determine

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

Expert Solution

Answer to Problem 23E

The equation of the tangent line to the equation x2xyy2=1 at (2,1) is y=3x412.

Explanation of Solution

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(1)]ddx(y2)

Apply the chain rule (1) and simplify the terms,

2x[xdydx+y(1)]=ddy(y2)dydx2xxdydxy=2ydydxdydx=y2xx2ydydx=2xyx+2y

Therefore, the derivative of y is dydx=2xyx+2y.

The slope of the tangent line at (2,1) is computed as follows,

m=dydx|(x,y)=(2,1)=2(2)12+2(1)=412+2=34

Thus, the slope of the tangent line at (2,1) is m=34.

Substitute (2,1) for (x1,y1) and m=34 in equation (1),

y1=34(x2)y1=3x464y=3x432+1y=3x412

Therefore, the equation of the tangent line to the equation x2xyy2=1 at (2,1) is y=3x412.

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