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Calculus 1 Problems & Solutions – Chapter 4 – Section 4.2.4

4.2.4
Logarithmic Differentiation

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Review

1. Differentiation Of The Function y = (f (x))^g⁽^x⁾

Example 1.1

Differentiate y = (2x)^sinx.

Solution 1
y = (2x)^sinx = e^sinxln²^x,

EOS

In Solution 2 below we take the natural logarithm of both sides of the given equation and differentiate implicitly both sides
of the resulting equation with respect to the variable, which here is x.

Solution 2
Using logarithmic differentiation we have:

ln y = ln (2x)^sinx = sin x ln 2x,

EOS

The given function is of the form ( f(x))^g⁽^x⁾, with f(x) = 2x and g(x) = sin x. The variable appears in both the base and
the exponent. Neither the power rule (d/dx) u^a = au^a^–1u' nor the exponent rule (d/dx) a^u = a^u(ln a)u' can be applied
directly in this case.

In Solution 1 we transform ( f(x))^g⁽^x⁾ into the exponential function using the definition u^v = e^v^lnu, to get ( f(x))^g⁽^x⁾ =
e^g⁽^x⁾^lnf⁽^x⁾. Then we differentiate e^g⁽^x⁾^lnf⁽^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⁾^lnf⁽^x⁾ back to ( f(x))^g⁽^x⁾.

In Solution 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, since
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 Solutions 1 and 2 yield the same answer.

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2. Differentiation Of Expressions Containing ( f (x))^g⁽^x⁾

Example 2.1

Find dy/dx if y = x³(sin x)^cosx.

Solution 1
y = x³(sin x)^cosx = x³e^cosxlnsinx,

EOS

Solution 2
Utilizing logarithmic differentiation we get:

ln y = ln x³ (sin x)^cosx = 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⁾^lnf⁽^x⁾ as in Solution 1 or logarithmic differentiation as in Solution 2. Again, in the answer don't
forget to replace e^g⁽^x⁾^lnf⁽^x⁾ by ( f(x))^g⁽^x⁾, or y by the expression of the given function.

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3. Differentiating Products And Quotients

Example 3.1

Find:

Solution
Let:

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.

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Problems & Solutions

1. Differentiate y = (sec x)^tanx in 2 ways:
a. Express it as natural exponential and then differentiate.
b. Use logarithmic differentiation.

Solution

a. y = (sec x)^{tan x} = e^{tanxlnsec x},
y' = e^tanxlnsec
x (sec² x ln sec x + tan x (1/secx) sec x tan x)
= (sec x)^{tan x} (sec² x ln sec x + tan² x).