Question

Find dy/dx in terms of x and y if x^2 + y^2 = sqrt(8).

I just don't know where to begin. Do you find the prime of x^2 + y^2 and then plug in dx/dy? I've been looking online and most people just do the math but don't explain what they are doing.

Answer 1

\[x^2+y^2=\sqrt{8}\]
\[2x+2yy'=0\] solve for \(y'\)

Answer 2

this is a version of "implicit differentiation"
you are thinking that \(y=f(x)\) even though you don't solve for \(y\) in terms of \(x\)

Answer 3

it is like taking the derivative of
\[x^2+f^2(x)=\sqrt{8}\] with respect to \(x\)
using the chain rule for the second part you get
\[2x+2f(x)f'(x)=0\] but it is easier to write
\[2x+2yy'=0\]

Answer 4

btw you can in fact solve for \(y\) in terms of \(x\) in this problem except you get a \(\pm\) when you do it
\[x^2+y^2=\sqrt{8}\]
\[y^2=\sqrt{8}-x^2\]
\[y=\pm\sqrt{x^2-\sqrt{8}}\]

Answer 5

Alright let me make sure I get this. (So I got the previous answer) which was -2x / (2y) now with this question:
x^3y - x - 7y - 8 = 0
what I would do is first find the derivative and my answer would be:
3x^2y-1
then I plug in y' after the y?
3x^2yy'-1
then I sold for y?
3x^2yy' = 1
y' = 1/(3x^2y)

Answer 6

\[x^3y - x - 7y - 8 = 0\]
\[x^3f(x)-x-7f(x)-8=0\] you need the product rule for the first one

Answer 7

\[x^3y-x-7y-8=0\]
\[3x^2y+x^3y'-1-7y'=0\] solve for \(y'\)