# 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 22E
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 curve sin(x+y)=2x2y at (π,π) is y=x3+2π3.

### Explanation of Solution

Given:

The curve is sin(x+y)=2x2y.

The point is (π,π).

Derivative rules: 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|(x1,y1).

Calculation:

Consider the equation sin(x+y)=2x2y.

Differentiate the above equation implicitly with respect to x,

sin(x+y)=2x2y

Differentiation with respect to x,

ddx(sin(x+y))=ddx(2x2y)ddx(sin(x+y))=ddx(2x)ddx(2y)ddx(sin(x+y))=2ddx(x)2ddx(y)ddx(sin(x+y))=22dydx

Let u=x+y and apply the chain rule,

ddx(sinu)=22dydxddu(sinu)dudx=22dydxcosududx=22dydx

Substitute u=x+y,

cos(x+y)ddx(x+y)=22dydxcos(x+y)[ddx(x)+dydx]=22dydxcos(x+y)[1+dydx]=22dydxcos(x+y)+cos(x+y)dydx=22dydx

Combine the terms dydx,

cos(x+y)dydx+2dydx=2cos(x+y)dydx=2cos(x+y)cos(x+y)+2

Therefore, the derivative of y is dydx=2cos(x+y)cos(x+y)+2.

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

m=dydx|(x,y)=(π,π)=2cos(π+π)cos(π+π)+2=2cos2πcos2π+2

Substitute the value cos2π=1,

m=211+2=13

Thus, the slope of the tangent line at (π,π) is m=13.

Substitute (π,π) for (x1,y1) and m=13 in equation (1),

yπ=13(xπ)yπ=x3π3y=x3π3+πy=x3+2π3

Therefore, the equation of the tangent line to the equation sin(x+y)=2x2y at (π,π) is y=x3+2π3.

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