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USE CODE: chait6

```				   how to find derative of sinx^ by first principle method
```

5 years ago

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### Answers : (3)

```										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.

Sagar Singh
B.Tech, IIT Delhi

```
5 years ago
```										limh->0 sin(x).
=limh->0sin(x+h)-sin(x).
=limh->0[sin(x)cos(h)+cos(x)sin(h)-sinx] / h.
=limh->0[sinx(cos(h)-1) +cos(x)sin(h)[/h.
=limh->0[Sinx(cos(h)-1)]/h + [cos(x)sin(h)]/h.
=0+cos(x). {(cosx-1)/x = 0. sinx/x=1}
=cos(x).
```
5 years ago
```										sinx^ what???
```
5 years ago

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