Question

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

Answer 1

It depends on what tools (theorems and definitions) you have to work with.

Explanation:

If we need to use the definition of derivative, we'll need the fundamental trigonometric limits:

`lim_(hrarr0)sinh/h=1` `" "` and `" "` `lim_(hrarr0)(1-cosh)/h=0`

along with `lim_(hrarr0)cos(x+h) = cosx`.

We'll also need the trigonometric identity:

`cos(x+h) = cosxcos h-sinxsin h`.

Here is the work.

For `f(x) = 1/cosx`, the definition of derivative gives

`f'(x) = (1/cos(x+h)-1/cosx)/h`.

As always, the initial form of the limit of this difference quotient is indeterminate. We need to work with the difference quotient until we get a limit of determinate form.

We'll start by getting a single fractional expression.

`(1/cos(x+h)-1/cosx)/h = ((1/cos(x+h)-1/cosx))/h * ((cos(x+h)cosx))/((cos(x+h)cosx))`

`= (cosx-cos(x+h))/(hcos(x+h)cosx)`

Now expand `cos(x+h)` and distribute the subtraction.

`= (cosx-cosxcos h+sinxsin h)/(hcos(x+h)cosx)`

Regroup so we can use thr fundamental trig limits.

`= (cosx(1-cos h)/h+sinxsin h/h)/(cos(x+h)cosx)`

Evaluate the limit by evaluating the individual limits

`lim_(hrarr0)(cosx(1-cos h)/h+sinxsin h/h)/(cos(x+h)cosx) = (cosx(0)+sinx(1))/(cosxcosx)`

`= sinx/cos^2x = 1/cosx sinx/cosx = secx tanx`

If you have the quotient rule

`d/dx(1/cosx) = (0cosx-1(-sinx))/(cosx)^2 = sinx/cos^2x`

Finish as above.

If you have the chain rule

Know that `d/dxf(x)^-1=-f(x)^-2*f'(x)`, so

`d/dx(cosx)^-1=-(cosx)^-2d/dxcosx=sinx/cos^2x`

And again finish as above.

If you know the derivative of secant , use `1/cosx = secx`