- #1
grzz
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Homework Statement
I was reading in a book that,
" For any function f(x) whose functional form changes from f(x) to f'(x), we can write
[itex]\bar{\delta}[/itex]f(x) = f'(x) - f(x)
= {f'(x') - f(x)} - {f'(x') - f'(x)}
= [itex]\delta[/itex]f(x) - [itex]\partial[/itex][itex]_{μ}[/itex]f(x)[itex]\delta[/itex]x[itex]^{μ}[/itex] "
I am understanding that the [itex]\bar{\delta}[/itex] represents only the change in the functional form at the same value of x and [itex]\delta[/itex] represents the total change.
My difficulty is in the term [itex]\partial[/itex][itex]_{μ}[/itex]f(x)[itex]\delta[/itex]x[itex]^{μ}[/itex].
Homework Equations
The Attempt at a Solution
If I expand {f'(x') - f'(x)} I get
f'(x[itex]^{μ}[/itex] + [itex]\delta[/itex]x[itex]^{μ}[/itex]) - f'(x[itex]^{μ}[/itex])
= f'(x[itex]^{μ}[/itex]) + [itex]\partial[/itex][itex]_{μ}[/itex]f'(x)[itex]\delta[/itex]x[itex]^{μ}[/itex] - f'(x[itex]^{μ}[/itex])
= [itex]\partial[/itex][itex]_{μ}[/itex]f'(x)[itex]\delta[/itex]x[itex]^{μ}[/itex] .
Hence I get the the [itex]\partial_{μ}[/itex] of f'(x) and not of f(x) as the book says.
Can somebody tell me what am I missing?