Question

How do you find the derivative of `y=1/cosx`?

Answer 1

`dy/dx = secx *tanx`

Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If `y=f(x)` then `f'(x)=dy/dx=dy/(du)(du)/dx`

I was taught to remember that the differential can be treated like a fraction and that the "`dx`'s" of a common variable will "cancel" (It is important to realise that `dy/dx` isn't a fraction but an operator that acts on a function, there is no such thing as "`dx`" or "`dy`" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

`dy/dx = dy/(dv)(dv)/(du)(du)/dx` etc, or `(dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx)`

So with `y = 1/cosx`, Then:

`{ ("Let "u=cosx, => , (du)/dx=-sinx), ("Then "y=1/u, =>, dy/(du)=-1/u^2 ) :}`

Using `dy/dx=(dy/(du))((du)/dx)` we get:

`\ \ \ \ \ dy/dx = (-1/u^2)(-sinx)`
`:. dy/dx = 1/cos^2x *sinx`
`:. dy/dx = 1/cosx *sinx/cosx`
`:. dy/dx = secx *tanx`