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

1 Answer
Aug 23, 2016

#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"#