Question

How do you differentiate `f(x)=4/cosx`?

Answer 1

`4tanxsecx`

Explanation:

differentiate using the `color(blue)"chain rule"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(a/a)|)))........ (A)`

`y=f(x)=4/(cosx)=4(cosx)^-1`

let `u = cosxrArr(du)/(dx)=-sinx`

and `y=4u^-1rArr(dy)/(du)=-4u^-2`

substitute these values into (A) convert u back into terms of x

`rArrdy/dx=-4u^-2(-sinx)=(4sinx)/(cos^2x)=(4sinx)/cosx xx1/cosx`

`color(orange)"Reminder"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(tanx=(sinx)/(cosx)" and " secx=1/(cosx))color(white)(a/a)|)))`

`rArrdy/dx=4tanxsecx`

Note that f(x) may also be differentiated using the `color(blue)"quotient rule"`