hw to find (d^2y/dx^2)
if
(2x^3-3y^2=8)

Question

hw to find (d^2y/dx^2)
if 
(2x^3-3y^2=8)

asnaseer · Answer

Use implicit differentiation. This video might help if you are unfamiliar with the technique: http://www.youtube.com/watch?v=5yTVUZCaU6k

anonymous · Answer

my answer is 2x

asnaseer · Answer

how did you get that - please show your steps

anonymous · Answer

$$2x^3-3y^2=$$

anonymous · Answer

$$6x^2-6y \frac{ dy }{ dx}=0$$
$$12x-6\frac{ d^2y }{dx^2 }=0$$
$$6\frac{ d^2y }{dx^2 }=12x$$
$$\frac{ d^2y }{ dx^2}=2x$$

asnaseer · Answer

the second term in the second step above is incorrect. you need to use the product rule here.

anonymous · Answer

how to use the product rule.. it mean udv+vdu>>> is this the formula.

asnaseer · Answer

yes

asnaseer · Answer

you might find it easier to write y' and y'' instead of dy/dx etc

anonymous · Answer

which part must be as u and v..

asnaseer · Answer

it doesn't matter - the rule is symmetric :)

asnaseer · Answer

d(uv)/dx = d(vu)/dx

amistre64 · Answer

might be easier to see the chain rule and product rule if y is written in function notation y(x)  just a thought.  It tends to make the equations bulkier, but easier to apply rules to in the learning process

asnaseer · Answer

-6y*y' is the term that you need to apply the product rule to. The -6 is just a constant multiplier so you can ignore that and just multiply the result by -6 at the end.

amistre64 · Answer

$$2x^3-3y^2=8$$
$$2x^3-3(y(x))^2=8$$
$$6x^2-6y(x)y'(x)=0$$
etc

anonymous · Answer

ok if i applied for  u=-6 v=y so the answer should be -6(dy/dx)+0----- using product rule.

asnaseer · Answer

no - the -6 is just a constant. as I said above, you need to apply it to y*y'.
so, for example, let u=y and let v=y'

asnaseer · Answer

whatever answer you get, multiply it all by -6 at the end

anonymous · Answer

ok..i thought that u=-6.. sorry for the mistake.

asnaseer · Answer

np :)

asnaseer · Answer

alternatively, you can take u=-6y and v=y'

anonymous · Answer

so now i get
$$12x-6(y \frac{ d^2y }{ dx^2 }+\frac{ d^2y }{ dx^2 }))$$
so can i solve this problem

asnaseer · Answer

still not correct.
try and do just the derivative of this first:$$-6y\frac{dy}{dx}$$

anonymous · Answer

wait first (dy/dx*dy/dx)=d^2y/dx^2

asnaseer · Answer

no - y' * y' = (y')^2

asnaseer · Answer

(y')^2 is not the same as y''

anonymous · Answer

ok.. for -6y(dy/dx)= 6y(y'')-6(y')^2

asnaseer · Answer

yes - that looks correct now.

asnaseer · Answer

wait - the initial term should also be negative

anonymous · Answer

sorry i fogot to put.
so how i going to solve it.

asnaseer · Answer

write out the final expression

anonymous · Answer

12x-6y(y'')-6(y')^2

asnaseer · Answer

you are missing the equals sign and the value on the right-hand-side

anonymous · Answer

12x-6y(y'')-6(y')^2=0

asnaseer · Answer

good - so first thing to spot is that you can divide through by 6

anonymous · Answer

2x-y(y'')-(y')^2=0 so how i going to get y''

asnaseer · Answer

now rearrange this to solve for y''.
you can substitute in the value for y' from your earlier step:$$6x^2-6y \frac{ dy }{ dx}=0$$

anonymous · Answer

y'=x^2/y

asnaseer · Answer

correct. use this in your equation involving y'' and then just rearrange to get y''

anonymous · Answer

final answer = y''=x^4/y^3-(2x/y)

asnaseer · Answer

doesn't look quite right

asnaseer · Answer

still not correct

anonymous · Answer

sorry y''=2x/y-(x^4/y^3)

asnaseer · Answer

nope

asnaseer · Answer

oops - sorry - misread - it IS correct

amistre64 · Answer

maybe easier to work thru this way, just an idea
$$D_x[2x^3-3y^2=8]$$
$$6x^2-6y~y'=0$$
$$-6y~y'=-6x^2$$
$$y'=\frac{x^2}{y}$$
$$y'=x^2y^{-1}$$
$$D_x[y'=x^2y^{-1}]=\frac{d^2y}{dx^2}$$

asnaseer · Answer

nice @amistre64 :)

anonymous · Answer

thank you very much..  all

asnaseer · Answer

yw :)