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Post a LessonAnswered on 06/04/2024 Learn Unit III: Calculus/Continuity and Differentiability
Sadika
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.
Answered on 06/04/2024 Learn Unit III: Calculus/Continuity and Differentiability
Sadika
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:
Let's check each condition:
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].
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).
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.
Answered on 06/04/2024 Learn Unit III: Calculus/Continuity and Differentiability
Sadika
To differentiate sin(2x)sin(2x) with respect to ecos(x)ecos(x), we can use the chain rule. Let u=ecos(x)u=ecos(x), then dudx=−esin(x)dxdu=−esin(x).
Now, we have:
ddu(sin(2x))=ddu(sin(2x))×dudxdud(sin(2x))=dud(sin(2x))×dxdu
Using the chain rule:
ddu(sin(2x))=dd(2x)(sin(2x))×d(2x)dxdud(sin(2x))=d(2x)d(sin(2x))×dxd(2x)
=cos(2x)×2=cos(2x)×2
=2cos(2x)=2cos(2x)
Now, we need to multiply this by dudxdxdu:
=2cos(2x)×(−esin(x))=2cos(2x)×(−esin(x))
=−2ecos(2x)sin(x)=−2ecos(2x)sin(x)
So, the derivative of sin(2x)sin(2x) with respect to ecos(x)ecos(x) is −2ecos(2x)sin(x)−2ecos(2x)sin(x).
Answered on 06/04/2024 Learn Unit III: Calculus/Continuity and Differentiability
Sadika
To verify Rolle's Theorem for the function f(x)=x2+2x−8f(x)=x2+2x−8 on the interval [−4,2][−4,2], we need to check three conditions:
Let's check these conditions one by one:
The function f(x)=x2+2x−8f(x)=x2+2x−8 is a polynomial function and is continuous everywhere. Therefore, it is continuous on the interval [−4,2][−4,2].
The function f(x)=x2+2x−8f(x)=x2+2x−8 is a polynomial function and is differentiable everywhere. Therefore, it is differentiable on the interval (−4,2)(−4,2).
Evaluate f(x)f(x) at the endpoints of the interval:
At x=−4x=−4: f(−4)=(−4)2+2(−4)−8=16−8−8=0f(−4)=(−4)2+2(−4)−8=16−8−8=0
At x=2x=2: f(2)=22+2(2)−8=4+4−8=0f(2)=22+2(2)−8=4+4−8=0
Since f(−4)=f(2)=0f(−4)=f(2)=0, Rolle's Theorem is satisfied on the interval [−4,2][−4,2].
Now, because f(x)f(x) is continuous on [−4,2][−4,2] and differentiable on (−4,2)(−4,2), and f(−4)=f(2)f(−4)=f(2), there must exist at least one point cc in the open interval (−4,2)(−4,2) such that f′(c)=0f′(c)=0, which is the conclusion of Rolle's Theorem.
Therefore, Rolle's Theorem is verified for the function f(x)=x2+2x−8f(x)=x2+2x−8 on the interval [−4,2][−4,2].
Answered on 06/04/2024 Learn Unit III: Calculus/Integrals
Sadika
To evaluate the integral
∫3axb2+c2x2 dx∫b2+c2x23axdx
we can use the substitution method. Let's make the substitution:
u=cxu=cx
Then, du=c dxdu=cdx, or dx=1c dudx=c1du.
Substituting uu and dxdx into the integral:
∫3axb2+c2x2 dx=∫3ab2+u2⋅1c du∫b2+c2x23axdx=∫b2+u23a⋅c1du
Now, we can factor out the constant 3acc3a and rewrite the integral in a more familiar form:
3ac∫1b2+u2 duc3a∫b2+u21du
This is a standard integral, known to be the arctangent function:
3ac⋅1barctan(ub)+Cc3a⋅b1arctan(bu)+C
Now, we need to substitute back u=cxu=cx:
3ac⋅1barctan(cxb)+Cc3a⋅b1arctan(bcx)+C
Thus, the integral evaluates to:
3abarctan(cxb)+Cb3aarctan(bcx)+C
where CC is the constant of integration.
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