Can h(x)=(cos x)^x be written as a composition of two functions f and

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The function h(x) = (cos x)^x cannot be expressed as a composition of two functions f and g, where f(x) = x^n and g(x) = cos x, due to the variable exponent x in h(x). The chain rule for differentiation does not apply in this case because the exponent in f(x) must be constant for the power law to hold. Attempts to redefine f(x) as a^x or f(x, n) = x^n still lead to inconsistencies with the variable exponent in h(x). The discussion emphasizes the importance of correct notation and understanding the implications of variable versus constant exponents in differentiation. Ultimately, defining f(x) to meet the requirements of h(x) proves impossible.
PeterPumpkin
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Can h(x)=(cos x)^x be written as a composition of two functions f and g where f(x)=x^n and g(x)=cosx ? where h=fog

REASON FOR ASKING: I am wondering this in connect with a differentiation I was having trouble with (but can now solve thanks to this forum). I mistakenly thought that I could apply the chain rule for composition of functions. Seems it doesn't apply. (https://www.physicsforums.com/showthread.php?p=2796762#post2796762)
 
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You need to be careful with notation. In your example
h(x)=f\circ g(x)=(\cos(x))^n
where you have a different variable in the exponent. So all rules you know are valid without exception, but you have to get the notation right.

You could also try
f(x)=a^x or f(x,n)=x^n
but you'll notice that at some point the expression won't match what you have in your rules.
 


It can't, because in h(x) the exponent is x, the variable, while in f(x) the exponent is a constant. This makes an important difference when you are differentiating because the standard "power law" only applies when the exponent is constant.
 


Gib Z said:
It can't, because in h(x) the exponent is x, the variable, while in f(x) the exponent is a constant. This makes an important difference when you are differentiating because the standard "power law" only applies when the exponent is constant.
Specifically, the derivative of xa, with a constant, is axa-1 while the derivative of ax is (ln(a))ax.
 


Thanks. I can see my mistake --- there's no way to define f(x) to satisfy the requirements.
 


There is. Remember that f(x) = e^{\ln f(x)}
 
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