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how to find derative of sinx^ by first principle method

```
9 years ago

```							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: All the best.
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Sagar Singh
B.Tech, IIT Delhi

```
9 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).
```
9 years ago
```							sinx^ what???
```
9 years ago
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