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To solve this problem, we will differentiate both sides of the given expression y=(tan⁡−1x)2y=(tan−1x)2 with respect to xx.

Let's denote y=(tan⁡−1x)2y=(tan−1x)2.

First, let's differentiate yy with respect to xx using the chain rule:

dydx=2tan⁡−1x⋅11+x2dxdy=2tan−1x⋅1+x21

Now, let's express dydxdxdy in terms of xx and yy:

dydx=2y⋅11+x2dxdy=2y⋅1+x21

Now, let's differentiate dydxdxdy with respect to xx:

d2ydx2=ddx(2y⋅11+x2)dx2d2y=dxd(2y⋅1+x21)

Using the product rule:

d2ydx2=2⋅11+x2⋅dydx+2y⋅(−2x(1+x2)2)dx2d2y=2⋅1+x21⋅dxdy+2y⋅(−(1+x2)22x)

Substituting dydx=2y⋅11+x2dxdy=2y⋅1+x21:

d2ydx2=2⋅11+x2⋅2y⋅11+x2+2y⋅(−2x(1+x2)2)dx2d2y=2⋅1+x21⋅2y⋅1+x21+2y⋅(−(1+x2)22x)

d2ydx2=4y(1+x2)2−4x2y(1+x2)2dx2d2y=(1+x2)24y−(1+x2)24x2y

d2ydx2=4y−4x2y(1+x2)2dx2d2y=(1+x2)24y−4x2y

d2ydx2=4(y−x2y)(1+x2)2dx2d2y=(1+x2)24(y−x2y)

Now, let's substitute y=(tan⁡−1x)2y=(tan−1x)2 back into the expression:

d2ydx2=4((tan⁡−1x)2−x2(tan⁡−1x)2)(1+x2)2dx2d2y=(1+x2)24((tan−1x)2−x2(tan−1x)2)

d2ydx2=4(tan⁡−1x)2(1−x2)(1+x2)2dx2d2y=(1+x2)24(tan−1x)2(1−x2)

Now, let's manipulate the expression d2ydx2dx2d2y to match the left side of the given expression (x2+1)2y2+2x(x2+1)y′=2(x2+1)2y2+2x(x2+1)y′=2:

(x2+1)2y2+2x(x2+1)y′=(x2+1)2((tan⁡−1x)2)2+2x(x2+1)dydx(x2+1)2y2+2x(x2+1)y′=(x2+1)2((tan−1x)2)2+2x(x2+1)dxdy

(x2+1)2y2+2x(x2+1)y′=(x2+1)2(tan⁡−1x)4+2x(x2+1)⋅2y1+x2(x2+1)2y2+2x(x2+1)y′=(x2+1)2(tan−1x)4+2x(x2+1)⋅1+x22y

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2(1+x2)2+4x(x2+1)y1+x2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2+1+x24x(x2+1)y

Now, let's simplify this expression:

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2+4x(x2+1)(tan⁡−1x)2(1+x2)(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)(tan−1x)2(1+x2)

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2+4x(x2+1)(tan⁡−1x)2(1+x2)(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)(tan−1x)2(1+x2)

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2+4x(x2+1)2(tan⁡−1x)2(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)2(tan−1x)2

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2(1+x2+x)(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2(1+x2+x)

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2(x2+1)(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2(x2+1)

(x2+1)2y2+2x(x2+1)y′=4(x2+1)(tan⁡−1x)2(x2+1)2y2+2x(x2+1)y′=4(x2+1)(tan−1x)2

(x2+1)2y2+2x(x2+1)y′=4(x2+1)y2(x2+1)2y2+2x(x2+1)y′=4(x2+1)y2

Thus, we have verified that (x2+1)2y2+2x(x2+1)y′=2(x2+1)2y2+2x(x2+1)y′=2.

To verify Rolle's Theorem for the function y=x2+2y=x2+2 on the interval [−2,2][−2,2], we need to check three conditions:

1. The function y=x2+2y=x2+2 is continuous on the closed interval [−2,2][−2,2].
2. The function y=x2+2y=x2+2 is differentiable on the open interval (−2,2)(−2,2).
3. The function takes the same value at the endpoints of the interval [−2,2][−2,2], i.e., y(−2)=y(2)y(−2)=y(2).

Let's check each condition:

1. Continuity: The function y=x2+2y=x2+2 is a polynomial function and is continuous everywhere. Therefore, it is continuous on the closed interval [−2,2][−2,2].

2. Differentiability: The function y=x2+2y=x2+2 is a polynomial function and is differentiable everywhere. Therefore, it is differentiable on the open interval (−2,2)(−2,2).

3. Endpoints: y(−2)=(−2)2+2=4+2=6y(−2)=(−2)2+2=4+2=6 y(2)=(2)2+2=4+2=6y(2)=(2)2+2=4+2=6

   So, y(−2)=y(2)y(−2)=y(2).

Since all three conditions of Rolle's Theorem are satisfied, there exists at least one cc in the open interval (−2,2)(−2,2) such that f′(c)=0f′(c)=0, where f(x)=x2+2f(x)=x2+2.

Now, let's find the derivative of y=x2+2y=x2+2:

y′=ddx(x2+2)=2xy′=dxd(x2+2)=2x

To find the critical point(s), we set y′=0y′=0: 2x=02x=0 x=0x=0

So, f′(0)=0f′(0)=0.

Thus, Rolle's Theorem is verified for the function y=x2+2y=x2+2 on the interval [−2,2][−2,2], and there exists at least one point cc in the open interval (−2,2)(−2,2) where the derivative f′(c)f′(c) is equal to zero.