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When we take differentiation of any function then what actually we do with that function? Ex.d/dx of x^2 is 2x. So what we have actually done with x^2.

Prasad Patil
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  • $(x^n)'=nx^{n-1}$, so $(x^2)'=2x^{2-1}=2x$.The differentation is the process of finding the derivative of the given function. See http://www.sosmath.com/calculus/diff/der00/der00.html. –  Aug 31 '15 at 17:57
  • Depends what you mean. The derivative function $f'(x)$ essentially a limiting process. But the differentiation operation can be thought as mapping a function to a derivative. – Karl Aug 31 '15 at 18:02

3 Answers3

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We take the limit $$\frac{d}{dx}\ x^2 = \lim_{\Delta x\to 0}\frac{(x+\Delta x)^2 - x^2}{\Delta x}$$ or generally $$\frac {d}{dx}\ f(x) = \lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

The right hand side is a function depending on $x$.

peterwhy
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  • Yes, but the limit is a number, not a function. Conceptually, whenever we have a particular value $x$, we take this limit, which will be different for every $x$. Now, construct (I repeat, in theory) a function that assigns that limit to every $x$: this is the derivative function. But this process cannot be implemented in practice, because it should be repeated for the infinite numbers $x$, so the rules of derivation provide a practical shortcut to the limit process. (This is basically a rephrasing of the comment by @Karl above) – Miguel Aug 31 '15 at 18:11
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    @MiguelAtencia But what I wrote is consistent with the definition of a function: you give me a particular $x$, I return you a number calculated from the limit (or I tell you the limit does not exist with that input $x$). – peterwhy Aug 31 '15 at 18:16
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Historically it was determining at each point $x$ the slope for $y(x) = x^2$ by using the slope $m$ of a secant of width $\Delta x$, running through the points $(x, y)$ and $(x+\Delta x, y + \Delta y)$ and then calculating the limit value of that slope $$ m = \frac{\Delta y}{\Delta x} = \frac{(x+\Delta x)^2-x^2}{\Delta x} = \frac{2x \Delta x+ (\Delta x)^2}{\Delta x} = 2x + \Delta x $$ when $\Delta x$ vanishes thus the secant turns into a tangent.

mvw
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The derivative of a continuous function $f(x)$ is defined as $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$

This is interesting to calculate, because it describes the slope of the tangent line to $f(x)$ in any given point $x$ if $f'(x)$ exists as a finite value. It is often viewed as the rate of growth for the function $f$.

In your example, we get that $f(x)=x^2$ has the derivative $$f'(x)=\lim_{h\rightarrow 0}\frac{(x+h)^2-x^2}{h}=\lim_{h\rightarrow 0}\frac{2xh+h^2}{h}=\lim_{h\rightarrow 0}(2x+h)=2x$$ which is the rate of growth for $f(x)=x^2$ in a given point $x$.

Scounged
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