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December 6, 2016
Posted by **Ed** on Sunday, October 2, 2011 at 8:40pm.

e^x/y^2= 8 + e^y; dy/dx

Please I need help and thank you very much!

- Calc -
**Count Iblis**, Sunday, October 2, 2011 at 9:10pme^x/y^2= 8 + e^y ---->

e^x = 8 y^2 + y^2 e^y

So, in this case, we can solve for x in terms of y. Then you can simply consider x to be a fnction of y, differentiate w.r.t. y, yielding dx/dy and dy /dx is then 1/(dx/dy).

But in general, you can't solve for either x or y. Then, if you have an equation of the form:

f(x,y) = 0

that defines the relation between x and y, you can find dy/dx by computing the differential of f in terms of dx and dy and equating that to zero.

The infinitesimal change due to a change in x by dx and a change in y by dy is:

df = (df/dx) dx + (df/dy) dy

where df/dx is the so-called partial derivative of the function w.r.t. x, which is the derivative of f w.r.t. x while considering y to be a constant. And df/dy is the partial derivative of f w.r.t. y, here you keep x constant.

Then, if x, and y satisfy the equation, we want x + dx and y + dy to also satisfy the equation. Since f(x,y) was zero and f(x + dx, y+dy) are zero, the change in f must be zero as well, so we put df = 0:

(df/dx) dx + (df/dy) dy = 0 ----->

dy/dx = - (df/dx)/(df/dy)