To determine

To find: The derivative dydx by implicit differentiation.

Answer

The differentiation of x2x+y=y2+1 is.dydx=x(x+2y)2y(x+y)2+x2.

Explanation

Given:

The equation x2x+y=y2+1.

Derivative rules:

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

dydx=dydu⋅dudx.

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

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]−f1(x)ddx[f2(x)][f2x]2.

Calculation:

Obtain the derivative of x2x+y=y2+1 implicitly with respect to x.

x2x+y=y2+1

Differentiate with respect to x on both sides,

ddx(x2x+y)=ddx(y2+1)(x+y)ddx(x2)−(x2)ddx(x+y)(x+y)2=ddx(y2+1)[(x+y)ddx(x2)]−[(x2)[ddx(x)+dydx]](x+y)2=ddx(y2)+ddx(1)[(x+y)2x]−[(x2)[1+dydx]](x+y)2=ddx(y2)+0

Apply the chain rule (1),

2x2+2xy−(x2+x2dydx)(x+y)2=[ddy(y2)dydx]2x2+2xy−x2(x+y)2−x2(x+y)2dydx=2ydydx2x2+2xy−x2(x+y)2=2ydydx+x2(x+y)2dydx2x2+2xy−x2(x+y)2=dydx(2y+x2(x+y)2)

Simplify further and obtain the derivative,

dydx(2y(x+y)2+x2(x+y)2)=x2+2xy(x+y)2dydx=x2+2xy(x+y)2×(x+y)22y(x+y)2+x2dydx=x(x+2y)2y(x+y)2+x2

Therefore, the differentiation of x2x+y=y2+1 is dydx=x(x+2y)2y(x+y)2+x2.