Question

What is the formula for implicit differentiatiion?

Answer 1

typically, you want to separate the y terms from the x terms.

Answer 2

same formula; just dont throw out your derived bits ...

Answer 3

and u cannot express y easily (or at all) in terms of x

Answer 4

d (3y^2) dy
------- = --- (6y)
dx dx

which is the same as usual ...

d (3x^2) dx
-------- = --- (6x)
dx dx


what you are used to doing tho is throwing out the dx/dx part isnce it equals 1

Answer 5

eg y^5 -y = x

Answer 6

an example would
be
find dy/dx when
3x^2 - 2xy = 4y^2 + x
you treat y as a function of x:
6x -( 2x*dy/dx + 2y) = 8y * dy/dx + 1
- you then make dy/dx the subject

note 2xy - use product rule
4y^2 - use of chain rule

Answer 7

Im still confused. But ill just guess on that question. lol

Answer 8

\[f(x,y)=0\quad,\quad y=y(x)\]
\[\frac{d}{dx}f(x,y)=0\]
\[\frac{\partial f}{\partial x}\cdot\frac{dx}{dx}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dx}=0\]
\[\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\cdot y'=0\quad\Rightarrow\quad y'=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}\]