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Higher order derivatives and higher order differentials |
Higher derivative formula for the product - Leibniz formula |
Higher derivatives of composite functions |
Higher derivatives of composite functions examples |
Higher derivatives of implicit functions |
Higher derivatives of implicit functions examples |
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Higher derivative formula for the product - Leibniz formula |
Recall
the product rule for the derivative of the product of two functions, |
(
u v )'
= u
v' + v
u' |
where,
u
and v
are functions of the variable x. |
By
applying the above rule we extract higher derivative formulas for the
product of two functions as follows, |
(
u v )''
=
(
u v'' + v'
u'
) + (
v u'' + u' v'
)
=
u
v'' + 2u' v'
+ u'' v |
further, (
u
v )'''
= u
v''' + 3u' v''
+ 3u'' v'
+ u''' v . |
Coefficients
in the above expressions are the same as in the binomial formula,
therefore |
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is
higher derivative formula for the product or Leibniz formula. |
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Higher derivatives of composite functions |
If
y = f (u) and
u
= g (x) such that f
is differentiable at u
and g
is differentiable at x
then,
we differentiate the composite
function y
[u (x)] by
applying the chain rule |
y'
(x)
= y'
(u) · u'
(x). |
To
get the second derivative of the given composite function we differentiate
the above expression using both the
product rule and the chain rule |
y''
(x)
= y'
(u) u''
(x) +
u'
(x) [y''
(u)
u'
(x)]
= y''
(u)
[u'
(x)] 2 +
y'
(u) u''
(x). |
Proceeding
the same way we get the third derivative of the composite function |
y'''
(x)
= y'''
(u)
[u'
(x)] 3
+
3y''
(u) u'
(x) u''
(x) +
y'
(u) u'''
(x) and so on. |
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Higher derivatives of composite functions
examples |
Example:
Find
the second derivative of |
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Solution: |
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Example:
Find
the second derivative of y
=
sin2x.
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Solution: |
y'
= 2sin x cos x,
y''
= 2[sin x( -sin
x)
+ cos
x cos x]
= 2(cos2x
-
sin2x)
= 2cos 2x. |
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Higher derivatives of implicit functions |
The
first derivative of a function y = f(x)
written implicitly as F(x,
y) = 0
is |
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In
words,
differentiate the implicit equation with respect to both variables at the
same time such that, when differentiating
with respect to x,
consider
y
constant, while
when differentiating with respect to y,
consider x constant
and multiply by y'. |
We
use this method to differentiate the
above equation to extract the second derivative |
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and
after substituting
the value for y' |
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Differentiating
the second
derivative equation above, applying the same method, we can get
the third derivative of
an implicit function and so on. |
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Higher derivatives of
implicit
functions examples |
Example:
Find
the second derivative of y2
=
2px
or written implicitly
y2
-
2px
=
0
as
F(x,
y) = 0.
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Solution: |
The
first derivative,
2yy'
= 2p
or
yy'
= p,
y'
= p/y. |
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The second derivative,
(y' )2
+ yy''
= 0 => y''
= -
(y' )2
/ y |
and
by substituting y'
= p/y,
y''
= -
p2
/ y3.
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We
can obtain the same result using the above formulas.
Therefore,
first calculate |
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and
plug these values into the formulas for the first and the second
derivative |
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Example:
Find
the second derivative of y
-
tan-1y
=
x.
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Solution:
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The second derivative, |
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Again,
we
can obtain the same result using the above formulas.
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