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OpenStudy (anonymous):

find d^2*y/dx^2 in terms of x and y for 1-xy=x-y?plzz:)..dont rlly understand what its asking:(

OpenStudy (anonymous):

this is asking for hte second derivative. So take the derivative of the derivative.

OpenStudy (anonymous):

\[d ^{2}y/dx ^{2}\]

OpenStudy (anonymous):

so first solve for y in terms of x.

OpenStudy (anonymous):

thts wat it says on the paper

OpenStudy (anonymous):

hm.

OpenStudy (anonymous):

are you sure:\[1-xy=x-y\] is the equation?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Implicit differentiation!

OpenStudy (anonymous):

exactly.

OpenStudy (anonymous):

only implicit diferntiation..thn wat does the d^2 n all mean?

OpenStudy (anonymous):

the squared means you do it tiwce. so you differentiate once, then do it again.

OpenStudy (anonymous):

n thn its over dx^2?..wat does tht mean?

OpenStudy (anonymous):

same thing, just done twice.

OpenStudy (anonymous):

ohh..ok..thank u soo much..rlly helps.. i wish thy jus wrote do the second deriv!

OpenStudy (anonymous):

basically when you with d^2(some function)/dx^2, you are differentiation the function in terms of x, then differentiating it again.

OpenStudy (anonymous):

\[\frac{d^2}{dx^2}=\frac{d}{dx}*\frac{d}{dx}\] Which means, effectively, take the second derivative. You can also take your function and solve it explicitly and avoid implicit completely. 1-xy=x-y 1-x=xy-y 1-x=(x-1)y y=(1-x)/(x-1) y=-(x-1)/(x-1) y=-1 the second derivative would be zero.

OpenStudy (anonymous):

Solve for x: -(x+xy)=-(y+1) x(y+1)=(y+1) x=1 The second derivative of that would also be zero.

OpenStudy (anonymous):

ohhok...understood:)! thank u so much!

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