fundamental rules of derivatives
fundamental rules of derivatives
Quotient Rule
MCV4U: Calculus & Vectors
Recap
Determine the derivative of f (x) =
3x 2 + 5
.
x −1
Using the quotient rule,
Implicit Differentiation
(x − 1)(6x) − (3x 2 + 5)(1)
(x − 1)2
2
3x − 6x − 5
=
(x − 1)2
f 0 (x) =
J. Garvin
Using the product and chain rules,
f (x) = (3x 2 + 5)(x − 1)−1
f 0 (x) = (6x)(x − 1)−1 − (3x 2 + 5)(x − 1)−2
which simplifies to the same after some work.
J. Garvin — Implicit Differentiation
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fundamental rules of derivatives
Implicit Differentation
fundamental rules of derivatives
Implicit Differentiation
Consider the line y = 3x + 5, whose derivative is
dy
dx
= 3.
Example
As it is written, the equation of the line is explicitly defined,
since it describes the line entirely in terms of x.
Determine the derivative of the line described by
3x − y = −5.
An equivalent equation for the line is 3x − y = −5. This
time, the equation is implicitly defined, since the function is
defined in terms of both x and y .
Take the derivative of both sides, noting that the right hand
side is a constant.
d
dx (3x − y )
d
d
3 dx
x − dx
y
While most functions that we have dealt with up to this
point have been explicitly defined, some fuctions are
impossible to express explicitly.
It is possible to determine the derivative of a function that is
implicitly defined by using the basic derivate rules that we
have established.
Note that
d
dx x
= 1 and
d
dx y
=
3−
fundamental rules of derivatives
Implicit Differentiation
dy
dx ,
dy
dx
so
=0
dy
dx
fundamental rules of derivatives
Implicit Differentiation
which is the derivative that we are looking for.
=3
This is the same answer that we receive when differentiating
y = 3x + 5.
In this instance, implicit differentiation is more work than
necessary, but there are other functions for which it is a
time-saver, or where it is the only option.
d
In the previous example, setting dx
y=
that the chain rule was being used.
dy
dx
obscured the fact
Consider the process of differentiating y = x 2 .
From the power rule, we know that
dy
dx
= 2x.
Alternatively, consider y as a composite function, where the
inner function is u = x and the outer function is y = u 2 .
Then
du
dx
= 1 and
dy
du
= 2u.
dy du
du · dx = 2u · 1 = 2x.
Note that when calculating dy
of
dx , we find the derivative
with respect to an “intermediate” variable u dy
,
then
du
multiply it by its derivative with respect to x du
dx .
Using the chain rule,
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=0
dy
dx ,
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J. Garvin — Implicit Differentiation
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Isolate
d
dx (−5)
=
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dy
dx
=
y
fundamental rules of derivatives
fundamental rules of derivatives
Implicit Differentiation
Implicit Differentiation
Example
Alternatively, we can implicitly differentiate x 2 + y 2 = 25
instead.
Determine the derivative of the circle described by
x 2 + y 2 = 25.
It is possible to define a circle explicitly using two equations,
one for the upper semicircle and one for the lower.
√
In this √
case, the equations would be y = 25 − x 2 and
y = − 25 − x 2 .
Differentiating the first equation using the chain rule results
1
2 − 21 (−2x), or dy = − √ x
in dy
.
dx = 2 (25 − x )
dx
25 − x 2
We could repeat the process for the second equation to
x
√
obtain dy
.
dx =
25 − x 2
2
2
d
dx (x + y )
d 2
d 2
dx x + dx y
d
dx 25
=0
d 2
x is simply 2x, since we are differentiating a
At this point, dx
function of x with respect to x.
d 2
dx y , however, is a function of y . So to differentiate this
term, use the chain rule.
2x + 2y dy
dx = 0
J. Garvin — Implicit Differentiation
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J. Garvin — Implicit Differentiation
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fundamental rules of derivatives
Implicit Differentiation
As before, isolate
=
dy
dx
Implicit Differentiation
to obtain the derivative.
dy
dx
fundamental rules of derivatives
Example
Determine the derivative of the hyperbola described by
3x 2 − 2y 2 = 4.
2x
2y
x
=−
y
=−
While possible to isolate y , implicit differentiation is faster.
2
d
dx (3x
Notice that in this case, the derivative is expressed in terms
of both x and y .
6x −
Therefore, if we wished evaluate the derivative of the
function, we would need to substitute values of x and y ,
rather than x alone.
√
Also notice, however, that since y = ± 25 − x 2 , then
x
dy
√
, which is the same result obtained earlier.
dx = ±
25 − x 2
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− 2y 2 ) =
4y dy
dx
dy
dx
d
dx 4
=0
6x
=
4y
3x
=
2y
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fundamental rules of derivatives
fundamental rules of derivatives
Implicit Differentiation
Implicit Differentiation
Example
The second term is the product of x and y , so we must use
the product rule here.
Determine the slope(s) of any tangent(s) to the ellipse
described by 2x 2 + 3xy + 5y 2 = 4 when x = 1.
d
4x + 3 dx
(xy ) + 10y dy
dx = 0
dy
4x + 3 (1)(y ) + (x) dy
+
10y
dx
dx = 0
It is not possible to isolate y in this relation, so explicit
differentiation is not an option.
2
2
d
dx (2x + 3xy + 5y )
d 2
d
d 2
2 dx x + 3 dx xy + 5 dx y
=
d
dx 4
=0
The first and third terms are not a problem, since we have
dealt with them before.
d
4x + 3 dx
(xy ) + 10y dy
dx = 0
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To isolate
dy
dx ,
dy
4x + 3y + 3x dy
dx + 10y dx = 0
we need to common factor.
dy
3x dy
dx + 10y dx = −(4x + 3y )
4x + 3y
dy
dx = − 3x + 10y
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fundamental rules of derivatives
Implicit Differentiation
fundamental rules of derivatives
Questions?
Now that we have an expression for the derivative, we need
to determine any points of tangency.
2(1)2 + 3(1)y + 5y 2 = 4
5y 2 + 3y − 2 = 0
(y + 1)(5y − 2) = 0
Thus, two points of tangency are (1, −1) and 1, 52 .
Substitute these values into the derivative to find the slopes.
4(1) + 3 52
dy 4(1) + 3(−1) dy =−
=−
2
dx (1,−1)
3(1) + 10(−1) dx (1, 2 )
3(1) + 10 5
5
= − 26
= 17
35
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J. Garvin — Implicit Differentiation
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