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Higher order derivatives and higher order differentials |
Higher derivatives of parametric functions |
Higher derivatives of parametric functions examples |
Higher order differentials |
Higher order differentials examples |
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Higher derivatives of parametric functions |
Assume
that f (t)
and g (t)
are differentiable and f
'(t) is not
0 then, given parametric curve can be
expressed as
y
= y (x) and this
function is differentiable at x,
that is |
x
= f (t)
or
t = f -1(x), |
by plugging
into y
= g (t)
obtained is y
= g [f -1(x)]. |
Therefore,
we use the chain rule and the derivative of the inverse function to find
the first derivative of parametric
functions, |
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The second
derivative of parametric
functions, |
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The
third derivative of parametric
functions, |
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Higher derivatives of parametric functions examples |
Example:
Find
the second derivative of the parametric functions x
= ln t and
y
= t3
+ 1.
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Solution:
Since |
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then, |
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Example:
Find
the second derivative of the parametric functions |
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Solution:
Since |
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and |
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then, |
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Higher order differentials |
The
first order differential of a function y
= f (x)
at a point x |
dy
= f '(x) dx = y'dx |
is
defined as the linear part of the increment of the function expressed as
the product of the derivative of the function
and the corresponding increment of the independent variable. |
The
second order differential is defined as the differential of the first
order differential with respect to the same increment dx,
written |
d
2 y
= d (dy) = d (y'dx) = y'' (dx)2. |
Similarly defined
are the third and the higher order differentials |
d
3 y
= y''' (dx)3, |
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d
n y
= y (n) (dx)n. |
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Furthermore,
if y
= f (u), where
u = g
(x),
then |
d
2 y
= y'' (du)2 +
y' d
2 u, |
d
3 y
= y''' (du)3 +
3 y''
du d 2 u
+
y' d
3 u,
etc. |
Note
that, here primes denote derivatives with respect to u. |
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Higher order differentials examples |
Example:
Find
d
2 y
of the function y
= cos 3x.
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Solution:
Therefore, given y
= f(u)
= cos
u,
where
u
= g(x)
= 3x.
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The
second differential we calculate using the formula |
d
2 y
= y'' (du)2 +
y' d
2 u,
where d
2 u
= g'' (x) (dx)2. |
Since,
y' (u)
=
-
3sin
3x
then, y''
(u)
=
-
9cos
3x |
and
g' (x)
= (3x)'
= 3
then, g''
(x)
= 0,
so that d
2 u
= 0. |
Thus, d
2 y
= -
9cos
3x
(du)2. |
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Example:
Find
d
2 y
of the
function y
= sin x
· ln x.
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Solution:
The second differential d
2 y
= y'' (dx)2,
since
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