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Mathematics 15 Online
OpenStudy (anonymous):

2 + q^2/200 = 4000/p^2 differentiate implicitly

OpenStudy (anonymous):

to find dq/dp

OpenStudy (anonymous):

yeh still not enough info , are we meant to assume that q ia a function of p

OpenStudy (anonymous):

sry...q is suppose demand for a good and p is the price of the good....

OpenStudy (anonymous):

yeh, so q is a function of p

OpenStudy (anonymous):

so it is the exact same as differentiating 2 + y^2 / 200 = 4000/x^2 where y=q=dependant variable

OpenStudy (anonymous):

\[2 + \frac{y^2}{200} = 4000x^{-2}\]

OpenStudy (anonymous):

diff wrt x ( and x=p ) 0 + (2y / 200 ) (dy/dx ) = -8000x^-3

OpenStudy (anonymous):

so dy/dx = -80000x^-3 / y

OpenStudy (anonymous):

\[\frac{dy}{dx} = \frac{-80000}{x^3 y}\]

OpenStudy (anonymous):

now replace y with q, and x with p

OpenStudy (anonymous):

ohhh ok thanks very much...

OpenStudy (anonymous):

wait, should be 800 000 on the top

OpenStudy (anonymous):

8000 x100 =800 000

OpenStudy (anonymous):

ok cool...i undersatnd your method though thats what i needed help with....

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