# The equation of the tangent line to the given equation at the point. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 3.5, Problem 24E
To determine

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

Expert Solution

The equation of the tangent line to the equation x2+2xy+4y2=12 at (2,1) is y=x2+2.

### Explanation of Solution

Given:

The curve is x2+2xy+4y2=12.

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 x2+2xy+4y2=12.

Differentiate the above equation implicitly with respect to x,

ddx(x2+2xy+4y2)=ddx(12)ddx(x2)+ddx(2xy)+ddx(4y2)=0ddx(x2)+2ddx(xy)+4ddx(y2)=0

Apply the product rule (2),

2x+2[xddx(y)+yddx(x)]+4ddx(y2)=02x+2[xdydx+y(1)]+4ddx(y2)=0

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

2x+2xdydx+2y+4[ddy(y2)dydx]=02x+2xdydx+2y+4[2ydydx]=02x+2xdydx+2y+8ydydx=0

Combine the terms dydx,

8ydydx+2xdydx=2x2ydydx(8y+2x)=2x2ydydx=2x+2y8y+2xdydx=x+y4y+x

Therefore, the derivative of the equation is dydx=x+y4y+x.

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

m=dydx|(x1,y1)=(2,1)=2+14(1)+2=36=12

Thus, the slope of the tangent line at (2,1) is m=12.

Substitute (2,1) for (x1,y1) and m=12 in equation (1),

y1=12(x2)y1=x2+1y=x2+1+1y=x2+2

Therefore, the equation of the tangent line to the equation x2+2xy+4y2=12 at (2,1) is y=x2+2.

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