Consider the function `y=x^(2)`. Calculate the derivative `(dy)/(dx)` using the concept of limit, at the point x=2.

Explanation

Let us take two points given by
`x_(1)=2andx_(2)=3`, then `y_(1)=4andy_(2)=9`
Here `Deltax=1andDeltay=5`,
Then `(Ø_(y))/(Ø_(x))=(9-4)/(3-2)=5`
If we take `x_(1)=2andx_(2)=2.5`, then `y_(1)=4andy_(2)=(2.5)^(2)=6.25`
Here `Deltax=0.5=(1)/(2)andDeltay=2.25`
Then `(Ø_(y))/(Ø_(x))=(6.25-4)/(0.5)=4.5`
If we take `x_(1)=2andx_(2)=2.25`, then `y_(1)=4andy_(2)=5.0625`.
Here `Deltax=0.25=(1)/(4),Deltay=1.0625`
`(Ø_(y))/(Ø_(x))=(5.0625-4)/(0.25)=((5.0625-4))/((1)/(4))=4(5.0625-4)=4.25`
If we take `x_(1)=2andx_(2)=2.1`, then `y_(1)=4andy_(2)=4.41`
Here `Deltax=0.1=(1)/(10)and(Ø_(y))/(Ø_(x))=((4.41-4))/((1)/(10))=10(4.41-4)=4.1`
These results are tabulated as shown below:


From the above table, the following inferences can be made.
As `Deltax` tends to zero, `(Deltay)/(Deltax)` approaches the limit given by the number 4.
At a point x=2, the derivative `(dy)/(dx)=4`.
It should also be mentioned here that `Deltaxto0` does not mean that `Deltax=0`. This is because, if we substitute `Deltax=0,(Deltay)/(Deltax)` becomes indeterminate.
In general, we can obtain the derivative of the function `y=x^(2)`, as follows:
`(Deltay)/(Deltax)=((x+Deltax)^(2))/(Deltax)=(x^(2)+2xDelta+Deltax^(2)-x^(2))/(Deltax)=(2xDelta+Deltax^(2))/(Deltax)=2x+Deltax`
`(dy)/(dx)=lim_(Deltaxto0)2x+Deltax=2x`