Question

Find the Derivative of `sec x` using first principle?

Answer 1

`d/dx sec x =tanx secx`

Explanation:

Define the function:

`f(x)=secx`

Using the limit definition of the derivative, we have:

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

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) (sec(x+h)-sec(x))/h`

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

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) ((cosx-cos(x+h))/(cos(x+h)cos(x)))/h`

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx-cos(x+h))/(hcos(x+h)cos(x))`

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx-(cosxcos h-sin x sin h))/(hcos(x+h)cos(x))`

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx-cosxcos h+sin x sin h)/(hcos(x+h)cos(x))`

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx(1-cos h)+sin x sin h)/(hcos(x+h)cos(x))`

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx(1-cos h))/(hcos(x+h)cos(x))+(sin x sin h)/(hcos(x+h)cos(x))`

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) ((1-cos h))/(hcos(x+h))+(tan x sin h)/(hcos(x+h))`

`\ \ \ \ \ \ \ \ = lim_(h rarr 0) (1-cos h)/h * sec(x+h)+(sin h)/h * tanx sec(x+h)`

Then we use two standard calculus limits:

`lim_(theta rarr 0) (1-cos theta)/theta = 0` and `lim_(theta rarr 0) (sin theta)/theta = 1`

Which gives us:

`f'(x) = 0 * sec(x) + 1 * tanx sec(x)`
`\ \ \ \ \ \ \ \ = tanx secx`