Calculus Of One Real Variable – By Pheng Kim Ving |
7.4 |
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1. Differentiation Of The Function y = ( f (x)) ^{g}^{ }^{(}^{x}^{)} |
Let's
differentiate the function y = ( f (x))^{g}^{(}^{x}^{)},
where both the base and the exponent are functions of x. Since the variable
appears in both the base and the exponent, we can't apply either the power rule
or the exponent rule to this function directly.
But first
we define ( f (x))^{g}^{(}^{x}^{)} in terms of f (x) and g(x). For x where f (x)
> 0, we define ( f (x))^{g}^{(}^{x}^{)} = e^{g}^{(}^{x}^{) }^{ln}^{ }^{f}^{ }^{(}^{x}^{)}. The expression
ln f (x) requires that f (x) > 0. That definition is similar to
that of b^{x}, which is b^{x}
= e^{x}^{ }^{ln}^{ }^{b}. For x where f (x)
= 0 and g(x) > 0,
we define ( f (x))^{g}^{(}^{x}^{)} = 0. Of course 0^{g}^{(}^{x}^{)} necessitates that g(x) > 0 for it to be defined.
Method 2
Taking
the natural logarithm of both sides of y = ( f (x))^{g}^{(}^{x}^{)} we have:
ln y = g(x) ln
f (x),
differentiate
both sides:
Logarithmic Differentiation
In
Method 1 we transform ( f (x))^{g}^{(}^{x}^{)} into the exponential function using the definition u^{v} = e^{v}^{ }^{ln}^{ }^{u}, to get ( f (x))^{g}^{(}^{x}^{)} =
e^{g}^{(}^{x}^{) }^{ln}^{ }^{f}^{ }^{(}^{x}^{)}. Then we differentiate e^{g}^{(}^{x}^{) }^{ln}^{ }^{f}^{ }^{(}^{x}^{)} with respect to x utilizing the exponent rule. This is
possible because the base e is a
constant. In the answer, we transform e^{g}^{(}^{x}^{) }^{ln}^{ }^{f}^{ }^{(}^{x}^{)} back to ( f (x))^{g}^{(}^{x}^{)}.
In
Method 2 we take the natural logarithm of both sides of the equation y = ( f (x))^{g}^{(}^{x}^{)}, to obtain ln y = ln ( f (x))^{g}^{(}^{x}^{)} =
g(x) ln f (x). Then we differentiate implicitly
both sides of the resulting equation ln y = g(x) ln f (x)
with respect to x. Note
that (d/dx) ln y = (1/y) dy/dx = y'/y, by the chain rule. Next we solve for y '. In
the answer, we replace y by ( f (x))^{g}^{(}^{x}^{)},
since we should express y '
in terms of x only, not of x and y. This technique is called logarithmic
differentiation, because
it involves the taking of the natural logarithm and the differentiation of the
resulting logarithmic equation. It allows us to
convert the differentiation of ( f (x))^{g}^{(}^{x}^{)} into the differentiation of a product.
Note
that both Methods 1 and 2 yield the same answer.
Remark 1.1
In
practice, you don't have to memorize the formula of the derivative of ( f (x))^{g}^{(}^{x}^{)}. That's welcome, because it's not easy
to
memorize it. Just remember one or both of the methods themselves.
Example 1.1
Differentiate
y = (2x)^{sin}^{ }^{x} in 2 different ways.
Solution 1
y = (2x)^{sin}^{ }^{x} = e^{sin}^{ }^{x}^{ }^{ln}^{ 2}^{x},
EOS
ln y = ln (2x)^{sin}^{ }^{x} = sin x ln 2x,
EOS
Of course in practice just use one method,
unless you're asked to use both methods.
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2. Differentiation Of Expressions Containing ( f (x)) ^{g}^{ }^{(}^{x}^{)} |
Example 2.1
Find dy/dx if y = x^{3}(sin x)^{cos}^{ }^{x} using 2 different methods.
Solution 1
y = x^{3}(sin x)^{cos}^{ }^{x} = x^{3}e^{cos}^{ }^{x}^{ }^{ln}^{ }^{sin}^{ }^{x},
EOS
ln y = ln x^{3} (sin x)^{cos}^{ }^{x} = 3 ln x + cos x ln sin x,
EOS
The
given function contains a term of the form ( f (x))^{g}^{(}^{x}^{)}, with f (x) = sin x and g(x) = cos x.
Hence we use either the
equation ( f (x))^{g}^{(}^{x}^{)} = e^{g}^{(}^{x}^{) }^{ln}^{ }^{f}^{ }^{(}^{x}^{)} as in Solution 1 or logarithmic differentiation as in
Solution 2. Again, in the answer don't forget to
replace e^{g}^{(}^{x}^{) }^{ln}^{ }^{f}^{ (}^{x}^{)} by ( f (x))^{g}^{(}^{x}^{)}, or y by the expression of the given
function.
Again
in practice just use one method, unless you're asked to
use both methods.
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3. Differentiating Products And Quotients |
Example 3.1
Find:
EOS
Here we
have a product and a quotient, but there's no term of the form ( f (x))^{g}^{(}^{x}^{)}, and we still employ logarithmic
differentiation, which therefore isn't exclusive for the form ( f (x))^{g}^{(}^{x}^{)}. Of course we can use the product and quotient rules, but
doing so would be more complicated. Generally, logarithmic differentiation is
advantageous when the products and/or
quotients are complicated. It enables us to convert the differentiation of a
product and that of a quotient into that of a sum and
that of a difference respectively.
Problems & Solutions |
1.
Differentiate y
= (sec x)^{tan}^{ }^{x} in 2 ways:
a. Express it as natural
exponential and then differentiate.
b. Use logarithmic
differentiation.
Solution
a.
y = (sec x)^{tan x}
= e^{tan}^{ }^{x}^{ }^{ln}^{ }^{sec x},
y' = e^{tan}^{ }^{x}^{ }^{ln}^{ }^{sec x} (sec^{2}^{ }x ln sec
x + tan x (1/sec
x) sec x tan
x)
= (sec x)^{tan x} (sec^{2}x ln sec
x + tan^{2} x).
b.
ln y = ln (sec
x)^{tan}^{ }^{x} = tan x ln sec x,
y' = y(sec^{2} x ln sec x + tan^{2}
x) = (sec x)^{tan}^{ }^{x} (sec^{2} x ln sec x + tan^{2}
x).
2.
Find y' using logarithmic differentiation if y = x^{x}/(x – 1)^{2}.
Solution
a.
Which of these functions grows more rapidly for sufficiently large x?
b. Differentiate them.
Solution
4.
Find:
Solution
Let:
5.
Find an equation of the tangent line to the curve:
at x = 0.
Solution
Consequently
the equation of the tangent line is y – 1 = 0(x – 0)
or y = 1.
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