Given function
We have
y = f ( x ) = sin 3 ( π 3 cos ( π 3 2 ( − 4 x 3 + 5 x 2 + 1 ) 3 2 ) ) . y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right). y = f ( x ) = sin 3 ( 3 π cos ( 3 2 π ( − 4 x 3 + 5 x 2 + 1 ) 2 3 ) ) . We need to evaluate the relation between y y y y ′ y' y ′ x = 1 x=1 x = 1
Rewrite in composite form
Let
u ( x ) = − 4 x 3 + 5 x 2 + 1 , u(x)=-4x^3+5x^2+1, u ( x ) = − 4 x 3 + 5 x 2 + 1 , so that
y = sin 3 ( π 3 cos ( π 3 2 ν ( x ) 3 / 2 ) ) . y=\sin^3\left(\frac{\pi}{3}\cos\left(\frac{\pi}{3\sqrt2}\,\nu(x)^{3/2}\right)\right). y = sin 3 ( 3 π cos ( 3 2 π ν ( x ) 3/2 ) ) . Now define
A ( x ) = π 3 2 ν ( x ) 3 / 2 , A(x)=\frac{\pi}{3\sqrt2}\,\nu(x)^{3/2}, A ( x ) = 3 2 π ν ( x ) 3/2 , B ( x ) = cos ( A ( x ) ) , B(x)=\cos(A(x)), B ( x ) = cos ( A ( x )) , C ( x ) = π 3 B ( x ) , C(x)=\frac{\pi}{3}B(x), C ( x ) = 3 π B ( x ) , so that
y = sin 3 ( C ( x ) ) . y=\sin^3(C(x)). y = sin 3 ( C ( x )) .
Compute values at x = 1 x=1 x = 1
First,
ν ( 1 ) = − 4 + 5 + 1 = 2. \nu(1)=-4+5+1=2. ν ( 1 ) = − 4 + 5 + 1 = 2. Hence
ν ( 1 ) 3 / 2 = 2 3 / 2 = 2 2 . \nu(1)^{3/2}=2^{3/2}=2\sqrt2. ν ( 1 ) 3/2 = 2 3/2 = 2 2 . Therefore
A ( 1 ) = π 3 2 ⋅ 2 2 = 2 π 3 . A(1)=\frac{\pi}{3\sqrt2}\cdot 2\sqrt2=\frac{2\pi}{3}. A ( 1 ) = 3 2 π ⋅ 2 2 = 3 2 π . So
B ( 1 ) = cos ( 2 π 3 ) = − 1 2 . B(1)=\cos\left(\frac{2\pi}{3}\right)=-\frac12. B ( 1 ) = cos ( 3 2 π ) = − 2 1 . Thus
C ( 1 ) = π 3 ( − 1 2 ) = − π 6 . C(1)=\frac{\pi}{3}\left(-\frac12\right)=-\frac{\pi}{6}. C ( 1 ) = 3 π ( − 2 1 ) = − 6 π . Therefore
y ( 1 ) = sin 3 ( − π 6 ) = ( − 1 2 ) 3 = − 1 8 . y(1)=\sin^3\left(-\frac{\pi}{6}\right)=\left(-\frac12\right)^3=-\frac18. y ( 1 ) = sin 3 ( − 6 π ) = ( − 2 1 ) 3 = − 8 1 .
Differentiate using chain rule
Since
y = sin 3 ( C ) , y=\sin^3(C), y = sin 3 ( C ) , we get
y ′ = 3 sin 2 ( C ) cos ( C ) C ′ . y'=3\sin^2(C)\cos(C)\,C'. y ′ = 3 sin 2 ( C ) cos ( C ) C ′ . Also,
C ′ = π 3 B ′ , B ′ = − sin ( A ) A ′ . C'=\frac{\pi}{3}B', \qquad B'=-\sin(A)A'. C ′ = 3 π B ′ , B ′ = − sin ( A ) A ′ . Hence
C ′ = − π 3 sin ( A ) A ′ . C'=-\frac{\pi}{3}\sin(A)A'. C ′ = − 3 π sin ( A ) A ′ . So
y ′ = 3 sin 2 ( C ) cos ( C ) ( − π 3 sin ( A ) A ′ ) . y'=3\sin^2(C)\cos(C)\left(-\frac{\pi}{3}\sin(A)A'\right). y ′ = 3 sin 2 ( C ) cos ( C ) ( − 3 π sin ( A ) A ′ ) . Thus
y ′ = − π sin 2 ( C ) cos ( C ) sin ( A ) A ′ . y'=-\pi\sin^2(C)\cos(C)\sin(A)A'. y ′ = − π sin 2 ( C ) cos ( C ) sin ( A ) A ′ .
Find A ′ ( x ) A'(x) A ′ ( x )
Recall
A ( x ) = π 3 2 ν ( x ) 3 / 2 . A(x)=\frac{\pi}{3\sqrt2}\,\nu(x)^{3/2}. A ( x ) = 3 2 π ν ( x ) 3/2 . So
A ′ ( x ) = π 3 2 ⋅ 3 2 ν ( x ) 1 / 2 ν ′ ( x ) . A'(x)=\frac{\pi}{3\sqrt2}\cdot \frac{3}{2}\nu(x)^{1/2}\nu'(x). A ′ ( x ) = 3 2 π ⋅ 2 3 ν ( x ) 1/2 ν ′ ( x ) . Now
ν ′ ( x ) = − 12 x 2 + 10 x . \nu'(x)=-12x^2+10x. ν ′ ( x ) = − 12 x 2 + 10 x . At x = 1 x=1 x = 1
ν ′ ( 1 ) = − 12 + 10 = − 2 , \nu'(1)=-12+10=-2, ν ′ ( 1 ) = − 12 + 10 = − 2 , ν ( 1 ) 1 / 2 = 2 . \nu(1)^{1/2}=\sqrt2. ν ( 1 ) 1/2 = 2 . Therefore
A ′ ( 1 ) = π 3 2 ⋅ 3 2 ⋅ 2 ⋅ ( − 2 ) = − π . A'(1)=\frac{\pi}{3\sqrt2}\cdot \frac{3}{2}\cdot \sqrt2\cdot (-2)=-\pi. A ′ ( 1 ) = 3 2 π ⋅ 2 3 ⋅ 2 ⋅ ( − 2 ) = − π .
Evaluate y ′ ( 1 ) y'(1) y ′ ( 1 )
At x = 1 x=1 x = 1
C ( 1 ) = − π 6 , A ( 1 ) = 2 π 3 . C(1)=-\frac{\pi}{6}, \qquad A(1)=\frac{2\pi}{3}. C ( 1 ) = − 6 π , A ( 1 ) = 3 2 π . Now,
sin ( − π 6 ) = − 1 2 , \sin\left(-\frac{\pi}{6}\right)=-\frac12, sin ( − 6 π ) = − 2 1 , sin 2 ( − π 6 ) = 1 4 , \sin^2\left(-\frac{\pi}{6}\right)=\frac14, sin 2 ( − 6 π ) = 4 1 , cos ( − π 6 ) = 3 2 , \cos\left(-\frac{\pi}{6}\right)=\frac{\sqrt3}{2}, cos ( − 6 π ) = 2 3 , sin ( 2 π 3 ) = 3 2 . \sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt3}{2}. sin ( 3 2 π ) = 2 3 . Thus
y ′ ( 1 ) = − π ( 1 4 ) ( 3 2 ) ( 3 2 ) ( − π ) . y'(1)=-\pi\left(\frac14\right)\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)(-\pi). y ′ ( 1 ) = − π ( 4 1 ) ( 2 3 ) ( 2 3 ) ( − π ) . Since
( 3 2 ) ( 3 2 ) = 3 4 , \left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)=\frac34, ( 2 3 ) ( 2 3 ) = 4 3 , we get
y ′ ( 1 ) = π 2 ⋅ 1 4 ⋅ 3 4 = 3 π 2 16 . y'(1)=\pi^2\cdot \frac14\cdot \frac34=\frac{3\pi^2}{16}. y ′ ( 1 ) = π 2 ⋅ 4 1 ⋅ 4 3 = 16 3 π 2 . So,
y ( 1 ) = − 1 8 , y ′ ( 1 ) = 3 π 2 16 . y(1)=-\frac18, \qquad y'(1)=\frac{3\pi^2}{16}. y ( 1 ) = − 8 1 , y ′ ( 1 ) = 16 3 π 2 .
Check each option
Option A
2 y ′ + 3 π 2 y 2y'+\sqrt3\pi^2 y 2 y ′ + 3 π 2 y At x = 1 x=1 x = 1
2 ( 3 π 2 16 ) + 3 π 2 ( − 1 8 ) = 3 π 2 8 − 3 π 2 8 ≠ 0. 2\left(\frac{3\pi^2}{16}\right)+\sqrt3\pi^2\left(-\frac18\right)
=\frac{3\pi^2}{8}-\frac{\sqrt3\pi^2}{8}
\neq 0. 2 ( 16 3 π 2 ) + 3 π 2 ( − 8 1 ) = 8 3 π 2 − 8 3 π 2 = 0. So A is false .
Option B
y ′ + 3 π 2 y y'+3\pi^2 y y ′ + 3 π 2 y At x = 1 x=1 x = 1
3 π 2 16 + 3 π 2 ( − 1 8 ) = 3 π 2 16 − 6 π 2 16 = − 3 π 2 16 ≠ 0. \frac{3\pi^2}{16}+3\pi^2\left(-\frac18\right)
=\frac{3\pi^2}{16}-\frac{6\pi^2}{16}
=-\frac{3\pi^2}{16}\neq 0. 16 3 π 2 + 3 π 2 ( − 8 1 ) = 16 3 π 2 − 16 6 π 2 = − 16 3 π 2 = 0. So B is false .
Option C
2 y ′ − 3 π 2 y \sqrt2 y'-3\pi^2 y 2 y ′ − 3 π 2 y At x = 1 x=1 x = 1
2 ⋅ 3 π 2 16 − 3 π 2 ( − 1 8 ) = 3 2 π 2 16 + 3 π 2 8 ≠ 0. \sqrt2\cdot \frac{3\pi^2}{16}-3\pi^2\left(-\frac18\right)
=\frac{3\sqrt2\pi^2}{16}+\frac{3\pi^2}{8}
\neq 0. 2 ⋅ 16 3 π 2 − 3 π 2 ( − 8 1 ) = 16 3 2 π 2 + 8 3 π 2 = 0. So C is false .
Option D
2 y ′ + 3 π 2 y 2y'+3\pi^2 y 2 y ′ + 3 π 2 y At x = 1 x=1 x = 1
2 ( 3 π 2 16 ) + 3 π 2 ( − 1 8 ) = 3 π 2 8 − 3 π 2 8 = 0. 2\left(\frac{3\pi^2}{16}\right)+3\pi^2\left(-\frac18\right)
=\frac{3\pi^2}{8}-\frac{3\pi^2}{8}=0. 2 ( 16 3 π 2 ) + 3 π 2 ( − 8 1 ) = 8 3 π 2 − 8 3 π 2 = 0. So D is true .
Final answer
The correct option is
D: 2 y ′ + 3 π 2 y = 0 . \boxed{\text{D: } 2y'+3\pi^2 y=0}. D: 2 y ′ + 3 π 2 y = 0 . This matches the stored correct answer.