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Transcript
Derivative Implicitly
Date:12/1/07
Long Zhao
Teacher:Ms.Delacruz
In calculus, a method called implicit
differentiation can be applied to
implicitly defined functions. This method is
an application of the chain rule allowing
one to calculate the derivative of a
function given implicitly.
As explained in the introduction, y can be
given as a function of x implicitly rather
than explicitly. When we have an equation
R(x,y) = 0, we may be able to solve it for
y and then differentiate. However,
sometimes it is simpler to differentiate
R(x,y) with respect to x and then solve for
dy / dx.
Source:Wikipedia
dy
Find
of
dx
x²+5x+3y²=21
Step 1:Take derivative on both sides
dy (x²)+dy(3y²)= dy (21)
dx
dx
dx
Step 2:Get the derivative
2x+6y dy =0
dx
dy
Step 3:Solve for
dy =-2x/6y
dx
dx
To find the equation of the
tangent line of
We find the derivative
The equation of the tangent line is:
y - 1 = 0.25 (x - 2)
Source:http://archives.math.utk.edu/v
isual.calculus/3/implicit.5/index.html
An example of an implicit function, for which
implicit differentiation might be easier than
attempting to use explicit differentiation, is
x³+2y²=8
In order to differentiate this explicitly, one would
have to obtain (via algebra)
Y=±
8 x
2
4
and then differentiate this function. This creates
two derivatives: one for y > 0 and another for
y < 0.
One might find it substantially easier to implicitly
differentiate the implicit function;
One might find it substantially easier to
implicitly differentiate the implicit
dy
function;4x³+4y =0
dx
thus,
dy  4 x 3  x 3


dx
4y
y
Source:wikipedia