Finally I've learned what a derivative is!
I've just started self-learning calculus.It's really interesting.It will be taught next year anyway.
I just want to solidify my understanding before officially studying it.
If I am asked to find the derivative of ##f(x)=x^2##.What I do is select two points and find the slope of the line passing throught two points.
$$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
$$=\frac{(x+\Delta x)^2-x^2}{\Delta x}$$
So the derivative of the graph at ##(x,f(x))## is the limit of the ##\frac{(x+\Delta x)^2-x^2}{\Delta x}## as ##\Delta x## approaches zero.
But if I make ##\Delta x## zero here,the fraction tuns to ##0 \over 0## :eek:
Then if I simplify it to ##\frac{\Delta x(2x+ \Delta x)}{\Delta x}##, I get 2x as the derivative.
Why? Isn't ##\frac{(x+\Delta x)^2-x^2}{\Delta x}=\frac{\Delta x(2x+ \Delta x)}{\Delta x}##
This is my problem.I don't have an intuitive understanding of this.
I've just started self-learning calculus.It's really interesting.It will be taught next year anyway.
I just want to solidify my understanding before officially studying it.
If I am asked to find the derivative of ##f(x)=x^2##.What I do is select two points and find the slope of the line passing throught two points.
$$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
$$=\frac{(x+\Delta x)^2-x^2}{\Delta x}$$
So the derivative of the graph at ##(x,f(x))## is the limit of the ##\frac{(x+\Delta x)^2-x^2}{\Delta x}## as ##\Delta x## approaches zero.
But if I make ##\Delta x## zero here,the fraction tuns to ##0 \over 0## :eek:
Then if I simplify it to ##\frac{\Delta x(2x+ \Delta x)}{\Delta x}##, I get 2x as the derivative.
Why? Isn't ##\frac{(x+\Delta x)^2-x^2}{\Delta x}=\frac{\Delta x(2x+ \Delta x)}{\Delta x}##
This is my problem.I don't have an intuitive understanding of this.
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