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∫ [f'(x)g(x) - g'(x)f(x)]/g(x)2 dx = f(x)/g(x) + C C is the constant of integration.
A composition function, regarding two functions, is when you apply the first function on the second function on an argument. Bear in mind that a single, unaltered function is when you apply said function to an argument; a composition function simply applies the result of an application as an argument to another function. For example, if one function is defined as f(x) = x + 4 and another is defined as g(x) = 2x, the composition of the two (where f is applied to g) is f(g(x)) = 2x + 4. Note that composition is not commutative; that is, f(g(x)) is not necessarily equivalent to g(f(x)), unless if the functions are either the same or inverses of each other, in which case the result will be the argument; f(f-1(x)) = f-1(f(x)) = x.
The chain rule states that the derivative of f(g(x)) is f'(g(x))⋅g'(x). In other words, it helps us differentiate *composite functions*. For example, sin(x²) is a composite function because it can be constructed as f(g(x)) for f(x)=sin(x) and g(x)=x².
Let f(x) = y y = 1 + (4/x) Now replace y with x and x with y and find equation for y x = 1 + (4/y) (x-1) = (4/y) y = 4/(x-1) This g(x), the inverse of f(x) g(x)= 4/(x-1) The domain will be all real numbers except when (x-1)=0 or x=1 So Domain = (-ā,1),(1,+ā) And Range = (-ā,0),(0,+ā) f(g(x)) = f(4/(x-1)) = 1 + 4/(4/(x-1)) = 1+(x-1) = x g(f(x)) = g(1+(4/x)) = 4/((1+(4/x))-1) = 4/(4/x) = x So we get f(g(x)) = g(f(x)) Notice the error in copying the next part of your question It should be g'(f(x)) = 1/(f'(g(x))) g'(f(x)) = d/dx (g(f(x))) = d/dx (x) = 1 f'(g(x)) = d/dx (f(g(x))) = d/dx (x) = 1 1/[f'(g(x))] = 1/1 = 1 g'(f(x)) = 1/f'(g(x)) ( Notice the error in copying your question)
Function "f" depends on "x", and function "g" depends on function "f".
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'Y' is a function 'f' of 'x': Y = f(x) . 'Z' is a function 'g' of 'y': Z = g [ f(x) ] .
Provided that the range of g(x) is the domain of f(x) then it is the composite function, called f of g of x.Note that f(g(x) ) is not the same as g(f(x).For example, if f(x) = x + 2 and g(x( = 3*x for real x, thenf(g(x)) = f(3*x) = 3*x + 2while g(f(x)) = g(x + 2) = 3*(x + 2) = 3*x + 6
∫ [f'(x)g(x) - g'(x)f(x)]/[f(x)g(x)] dx = ln(f(x)/g(x)) + C C is the constant of integration.
if f(x) = 4x, then the inverse function g(x) = x/4
∫ [f'(x)g(x) - g'(x)f(x)]/g(x)2 dx = f(x)/g(x) + C C is the constant of integration.
You would have been given a function for f(x) and another function for g(x). When you want to find f(g(x)), you put the function for g(x) wherever x occurs in f(x). Example: f(x)=3x+2 g(x)=x^2 f(g(x))=3(x^2)+2 I'm not sure what you mean by address domain and range. They depend on what functions you're given.
∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g/(x) dx This is known as integration by parts.
A composition function, regarding two functions, is when you apply the first function on the second function on an argument. Bear in mind that a single, unaltered function is when you apply said function to an argument; a composition function simply applies the result of an application as an argument to another function. For example, if one function is defined as f(x) = x + 4 and another is defined as g(x) = 2x, the composition of the two (where f is applied to g) is f(g(x)) = 2x + 4. Note that composition is not commutative; that is, f(g(x)) is not necessarily equivalent to g(f(x)), unless if the functions are either the same or inverses of each other, in which case the result will be the argument; f(f-1(x)) = f-1(f(x)) = x.
∫ [1/[f(x)(f(x) ± g(x))]] dx = ±∫1/[f(x)g(x)] dx ± (-1)∫ [1/[g(x)(f(x) ± g(x))]] dx