derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Derivatives Involving Exponential/Logarithmic
Functions
MCV4U: Calculus & Vectors
Recap
Determine the derivative of f (x) = 3x log4 (5x 2 − 3x).
Use the product and chain rules here.
Logarithmic Differentiation
f 0 (x) = 3x ln 3 · log4 (5x 2 − 3x) + 3x ·
J. Garvin
10x−3
(5x 2 −3x) ln 4
J. Garvin — Logarithmic Differentiation
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Slide 1/11
derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Logarithmic Differentiation
Logarithmic Differentiation
We have covered several derivative rules so far (e.g. power
rule, product rule, chain rule), as well as implicit
differentiation.
Example
Logarithmic differentiation is a technique that introduces
logarithms into a function in order to rewrite it in a
differentiable way.
In this case, y is neither a polynomial (the exponent is not a
constant) nor an exponential function (the base is not a
constant), so neither rule applies.
This is usually done by using the property logb p q = q logb p,
“bringing down” the exponent as a product, then using
implicit differentiation to find the derivative.
By taking the natural logarithm of both sides, however, we
are able to “pull down” the exponent as a constant multiple
using logarithm rules.
Many functions require the use of logarithmic differentiation
to find their derivatives.
ln y = ln x x
Determine the derivative of y = x x .
ln y = x ln x
Since the natural logarithm is generally easier to work with
than a general logarithm, most solutions involve ln.
J. Garvin — Logarithmic Differentiation
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J. Garvin — Logarithmic Differentiation
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derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Logarithmic Differentiation
Logarithmic Differentiation
Now we can use implicit differentiation. Be sure to use the
product rule on the right hand side.
Example
1 dy
y dx
= ln x + x ·
1
x
= ln x + 1
dy
dx
= y (ln x + 1)
= x x (ln x + 1)
Determine the derivative of y = (sin x)2x .
As before, take the natural logarithm of both sides and bring
down the exponent.
ln y = 2x ln(sin x)
1 dy
y dx
dy
dx
cos x
sin x
x
2x · cos
sin x
= 2 ln(sin x) + 2x ·
= y ·2 ln(sin x) +
= (sin x)2x · 2 ln(sin x) + 2x ·
J. Garvin — Logarithmic Differentiation
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J. Garvin — Logarithmic Differentiation
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cos x
sin x
derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Logarithmic Differentiation
Logarithmic Differentiation
Logarithmic differentiation can be used with any function,
but would not generally make sense for simple functions (e.g.
polynomials), since it introuduces both implicit differentiation
and the chain rule.
Example
On the other hand, logarithmic differentiation can be used as
an alternative when functions are significantly more complex.
We can, of course, use the product, quotient and chain rules
to find dy
dx .
√
√
3x 2 x 2 + 1(1 − 2x) − x 3 −2 x 2 + 1 + (1 − 2x) 2√2x
2
dy
x +1
=
dx
(x 2 + 1)(1 − 2x)2
In most cases, logarithmic differentiation will involve
logarithm laws to either simplify the derivative, or to make
the process more straightforward (though not necessarily
shorter).
Determine the derivative of y = √
x2
x3
.
+ 1(1 − 2x)
This is pretty intimidating, and difficult to simplify. Instead,
repeat the question using logarithmic differentiation.
J. Garvin — Logarithmic Differentiation
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J. Garvin — Logarithmic Differentiation
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derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Logarithmic Differentiation
Logarithmic Differentiation
Begin by taking the natural logarithm of both sides.
Now use implicit differentiation, along with the chain rule.
y=√
ln y = ln
x3
x2
+ 1(1 − 2x)
x3
√
x 2 + 1(1 − 2x)
Now,use
the properties logb (pq) = logb p + logb q and
logb pq = logb p − logb q.
p
ln y = ln x 3 − ln( x 2 + 1(1 − 2x))
p
= ln x 3 − ln x 2 + 1 − ln(1 − 2x)
J. Garvin — Logarithmic Differentiation
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derivatives of trigonometric, exponential & logarithmic functions
Questions?
J. Garvin — Logarithmic Differentiation
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2x
1 dy
y dx
dy
dx
√
2
3x 2
−2
− √2 x +1 −
x3
x 2 + 1 1 − 2x
3
2x
2
= −
+
x
2(x 2 + 1) 1 − 2x
3
2x
2
=y
−
+
x
2(x 2 + 1) 1 − 2x
x3
3
2x
2
=√
−
+
2
2
2(x + 1) 1 − 2x
x + 1(1 − 2x) x
=
While the answer itself is not a whole lot “simpler”, the
process is arguably easier – finding the derivative of the three
terms involving ln uses only the chain rule.
J. Garvin — Logarithmic Differentiation
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