Differentiation and derivative

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Differentiation and derivative

Mathematics
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Could you kindly walk me through the significance of each? I haven't done calculous in a while now .
I am not sure as to the significance of f besides the bracket (x+deltax) please excuse my grave ignorance.

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\[y+Deltay=f(x+Deltax)=m(x+Deltax)+b=mx+m Deltax+b=y+m Deltax\]
Informal definition: \(f\) denotes a function, which is essentially a rule that matches/maps a member of one set (the domain) with a single member of another set (the codomain, or range). (Note that this allows for several members from the domain to be matched with the same member from the range.) Example: Suppose you have this machine that, upon the press of a button, has the amazing ability to transform a dog into a cat. You can think of this capacity to alter the species as a function. Any given dog that enters the contraption belongs to the function's domain, while any cat that exits it belongs to the codomain. In this scenario, for every dog that enters the machine, only one cat comes out. Symbolically, we can denote this species-alteration by the letter \(f\), followed by a pair of brackets. You can think of \(f\) as the button, and the brackets as the door of the machine. \[f(\text{dog})=\text{cat}\]
and the equal sign as the exit platform
I had another part written up but I lost the connection. Here it is: You have this function \(y=f(x)\). This notation is used to mean that \(y\) is some value that depends on the value of \(x\). The function \(f\) is the matching rule that assigns any given value of \(x\) (so long as it's in the domain) to a particular value of \(y\). The \(\Delta\) symbol is traditionally used to denote the notion of change. (\(\Delta\) is the Greek uppercase letter \(D\); think "D" for "difference".) For instance, if \(y=f(x)\) denotes the temperature \(y\) as a function of time \(x\), then \(\Delta x\) can be thought of as the change in time, i.e. given two specified instants of time \(x_1\) and \(x_2\) (where \(x_2>x_1\)), you have \(\Delta x=x_2-x_1\). (The condition that \(x_2>x_1\) isn't necessary, but it's often useful to use this definition for \(\Delta x\), called the "forward difference".) Since \(y\) is a function of \(x\), then it's reasonable to believe that any change that occurs in \(x\) will result in some change to occur in \(y\). So we have some function \(y=f(x)\) that's sensitive to the value of the independent variable \(x\). If \(x\) changes, i.e. we add some amount \(\Delta x\) to \(x\), then we have some new value \(x+\Delta x\). Now if we apply our function, we have \(f(x+\Delta x)\). We don't how this affects \(y\) exactly unless we have the specific information regarding \(f\) and the amount of change \(\Delta x\), but we can predict \(y\) to change appropriately by some amount \(\Delta y\). So, you end up with what you're given in that image: \[y+\Delta y=f(x+\Delta x)\]
According to your image, the function happens to be linear, i.e. of the form \(y=f(\color{red}x)=m\color{red}x+b\). If we replace \(x\) with \(x+\Delta x\), we get \[y+\Delta y=f(\color{red}{x+\Delta x})=m(\color{red}{x+\Delta x})+b\] The rest of the simplification follows naturally.

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