maths >> differentiation?

Question

Q: If y = ( cos x )^1/2 , show that 4y^3(d2y/dx2) + y^4 +1 = 0

Sorry that just now post the wrong question. This is the correct question, anyone can guide?

but i'm sure that this time i copy the correct question.

i try the question by movind the power 1/2 to lest side, which means i make the y become y^2 before i start diff. w.r.t x. Then i straight away do the whole things until d2y/dx2. i do sub. at last to change it into the form that require. Can show me "d2y/dx2" of y = ( cos x )^1/2 ?

Finally i have figure it out but not sure whether is it acceptable or not..

y = ( cos x )^1/2
y^2 = cos x

diff. w.r.t x,
2y(dy/dx) = - sin x

diff. w.r.t x,
2y(d2y/dx2) + 2(dy/dx)^2 = - cos x

2y(d2y/dx2) + 2[( - sin x ) / ( 2y )]^2 = -y^2 (since y^2 = cos x, so i sub. it here)

2y(d2y/dx2) + [( sin^2 (x) ) / ( 2y^2 )] = -y^2

4y^3(d2y/dx2) + sin^2 (x) = -2y^4

Since y^2 = cos (x), so,
y^4 = cos^2 (x)

And using trigonometry identity,,
sin^2 (x) + cos^2 (x) = 1
sin^2 (x) = 1 - cos^2 (x)
= 1 - y^4


So,
4y^3(d2y/dx2) + 1 - y^4 = -2y^4
4y^3(d2y/dx2) + y^4 + 1 = 0

So it's proven. Can be accept or not ?

Answer 1

OH, you wanted to get that second equation using other methods.....

The way you put the question made it seem like you wanted us to prove that it was EQUAL TO 0.

Sorry, we all misunderstood.

Anyway your method is right, and next time make your question more clear.

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Answer 2

y=[cosx]^1/2, hence dy/dx= 1/2.[sin]^12,
d2y/dx2 = -1/4.[cosx]^-3/2. 4y^3 = 4[cosx]^3/2
y^4= [cosx]^2 adding terms gives

4[cosx]^3/2. -1/4.[cosx]^-3/2 + [cosx]^2 + 1 =
-1 + [cosx]^2 + 1 = [cosx]^2 ?
not 0 ? unless I have made a mistake !

Answer 3

I think you still copied it down wrong, because i get

subs(y=cos(x)^(1/2),4*y^3*diff(y,x$2)+y^4+1);
1 + cos(x)^2