# Thread: Second Derivative of Trigonometry function

1. ## Second Derivative of Trigonometry function

Dear All,

I'm a newbie here and wish to seek assistance from expert here.

I have one question in the past year exam question. the question is

(4y^3)*(d^2y/dx^2)+(y^4)+1=0

where y=(cos x)^1/2

i really seek assistance from experts here..

thank you,

2. ## Re: Second Derivative of Trigonometry function

Originally Posted by wa1515
Dear All,

I'm a newbie here and wish to seek assistance from expert here.

I have one question in the past year exam question. the question is

(4y^3)*(d^2y/dx^2)+(y^4)+1=0

where y=(cos x)^1/2

i really seek assistance from experts here..

thank you,
$y'_x=\frac{1}{2\sqrt{\cos x}}\cdot (-\sin x)=\frac{-\sin x}{2\sqrt{\cos x}}$

$y''_x=\frac{(-\sin x)'_x\cdot 2\sqrt{\cos x}-(-\sin x)\cdot(2\sqrt{\cos x})'_x}{4\cos x}$

3. ## Re: Second Derivative of Trigonometry function

Originally Posted by wa1515
Dear All,

I'm a newbie here and wish to seek assistance from expert here.

I have one question in the past year exam question. the question is

(4y^3)*(d^2y/dx^2)+(y^4)+1=0

where y=(cos x)^1/2

i really seek assistance from experts here..

thank you,
It would help if you posted the actual question. Are we to assume that the actual question asks you to verify y=(cos x)^1/2 as a solution to the DE and your problem is in calculating d^2y/dx^2?