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Usually very little.

f(x) simply denotes the function f except that with this notation it is explicit that f(x) is a function of the variable x (and only x).

So, f = x + 3 and f(x) = x + 3 are equivalent.

However, there will be times, with functions of more than one variables where you wish to consider only one of the arguments while holding the other argument constant.

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Looks like simple difference formula work here. f(x) - g(x) = f'(x) - g'(x) e - 5x = e - 5 =====


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Backward difference is a technique used in numerical analysis for approximating derivatives. For instance, if you have a function ( f(x) ) and you want to approximate the first derivative at a point ( x_0 ), you can use the backward difference formula: [ f'(x_0) \approx \frac{f(x_0) - f(x_0 - h)}{h} ] where ( h ) is a small step size. An example would be if ( f(x) = x^2 ), then the backward difference at ( x_0 = 2 ) with ( h = 0.1 ) would yield ( f'(2) \approx \frac{4 - 3.61}{0.1} = 3.9 ), which approximates the true derivative ( f'(x) = 2x ) at ( x = 2 ).


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What is the derivative of e-5x?

Looks like simple difference formula work here. f(x) - g(x) = f'(x) - g'(x) e - 5x = e - 5 =====


Solved examples of backward difference questions under numerical analysis?

Backward difference is a technique used in numerical analysis for approximating derivatives. For instance, if you have a function ( f(x) ) and you want to approximate the first derivative at a point ( x_0 ), you can use the backward difference formula: [ f'(x_0) \approx \frac{f(x_0) - f(x_0 - h)}{h} ] where ( h ) is a small step size. An example would be if ( f(x) = x^2 ), then the backward difference at ( x_0 = 2 ) with ( h = 0.1 ) would yield ( f'(2) \approx \frac{4 - 3.61}{0.1} = 3.9 ), which approximates the true derivative ( f'(x) = 2x ) at ( x = 2 ).