# Use first principles to find the derivative of f(x) = sec x

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### 1 Answer

The derivative of a function f(x) is defined as `lim_(h->0) (f(x+h) - f(x))/h` .

For f(x) = sec x, the derivative from first principles is:

`f'(x) = lim_(h->0)(sec(x+h) - sec x)/h`

`= lim_(h->0)(1/(cos(x+h)) - 1/(cos x))/h`

`= lim_(h->0) (cos x - cos(x+h))/(cos(x+h)*cos x*h)`

` ` = `lim_(h->0)(cos x - (cos x*cos h - sin x*sin h))/(cos(x+h)*cos x*h)`

`= lim_(h->0) (cos x(1 - cos h))/(cos(x+h)*cos x*h) + (sin x*sin h)/(cos(x+h)*cos x*h)`

`= lim_(h->0) (1 - cos h)/(cos(x+h)*h) + (sin x/cos x)*lim_(h->0) (sin h)/(cos(x+h)*h)`

At h = 0, sin h = 0 and cos h = 1. Also,` lim_(h->0)(1 - cos h)/h = 0` and `lim_(h->0) (sin h / h) = 1`

Using the relations given earlier, the derivative is:

`0 + (sin x/cos x)*1/(cos x)`

= sec x*tan x

**The derivative of f(x) = sec x is f'(x) = sec x*tan x**

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