   Chapter 3.5, Problem 16E ### 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

# Find dy/dx by implicit differentiation.16. x y = x 2 + y 2

To determine

To find: The derivative dydx by implicit differentiation.

Explanation

Given:

The equation xy=x2+y2.

Derivative rules:

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

dydx=dydududx.

(2) Product Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f(x)g(x)]=f(x)ddx(g(x))+g(x)ddxf(x).

Calculation:

Obtain the derivative of xy=x2+y2 implicitly with respect to x.

xy=x2+y2

Differentiate with respect to x on both sides.

ddx(xy)=ddx(x2+y2)

Let u=x2+y2 and apply the chain rule (1).

ddx(xy)=ddx(u)ddx(xy)=ddx(u12)

Apply the product rule (2) and the chain rule (1).

xddx(y)+yddx(x)=ddu(u12)dudxxdydx+y(1)=(12u121)dudxxdydx+y=(12u12)dudx

Substitute the value u=x2+y2 and simplify the expression,

xdydx+y=(12(x2+y2)12)ddx((x2+y2))xdydx+y=(12(x2+y2)12)[ddx(x2)+ddx(y2)]xdydx+y=(12(x2+y2)12)[2x

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