Dear vishal

We can approximate this value by taking a point somewhere near to P(x, f(x)), say Q(x + h, f(x + h)).

The value is an approximation to the slope of the tangent which we require.

We can also write this slope as "change in y / change in x" or:

If we move Q closer and closer to P, the line PQ will get closer and closer to the tangent at P and so the slope of PQ gets closer to the slope that we want.

If we let Q go all the way to touch P (i.e. h = 0), then we would have the exact slope of the tangent.

Now, can be written:

So also, the slope PQ will be given by:

But we require the slope at P, so we let h → 0 (that is let h approach 0), then in effect, Q will approach P and will approach the required slope.

Putting this together, we can write the slope of the tangent at P as:

This is called differentiation from first principles, (or the delta method). It gives the instantaneous rate of change of y with respect to x.

This is equivalent to the following (where before we were using h for Δx):

You will also come across the following for delta method:

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

 

Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

lim_h->0 sin(x).

=lim_h->0sin(x+h)-sin(x).

=lim_h->0[sin(x)cos(h)+cos(x)sin(h)-sinx] / h.

=lim_h->0[sinx(cos(h)-1) +cos(x)sin(h)[/h.

=lim_h->0[Sinx(cos(h)-1)]/h + [cos(x)sin(h)]/h.

=0+cos(x). {(cosx-1)/x = 0. sinx/x=1}

=cos(x).

sinx^ what???