What is the derivative of #sec(x-x^2)#?

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 = sec(x-x^2) #, Then:

# { ("Let "u=, => , (du)/dx=1-2x), ("Then "y=secu, =>, dy/(du)=secutanu ) :}#

(NB you should know that #color(blue)(d/dx secx=secxtanx) # ; If you don't then learn it!)

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

# dy/dx = secutanu(1-2x) #
# dy/dx = (1-2x) sec(x-x^2)tan(x-x^2)#

you start of with:
#sec(x-x^2)#

The derivative of sec(x) is sex(x)tan(x) because:

USELESS UNLESS YOU WANT TO KNOW HOW TO GET THE DERIVATIVE SEC(X) IF YOU FORGOT THE FORMULA:

sec(x)=#1/cos(x)#

since we have #1/cos(x)#, we will use the quotient rule, which states:

#g/h# = #(g' h - h' g) / g^2 #

the derivative of cos(x) is -sin(x), and the derivative of 1 is 0:

#(cos(x) (0) - (1) -sin(x))/ cos(x)^2#

#sin(x)/cos(x)^2#

#1/cos(x) * sin(x)/cos(x)#

#1/cos(x)# = sec(x) and #sin(x)/cos(x)# = tan(x)

Continuing on with the previous discussion
sec(x) is sex(x)tan(x), so:
#sec(x-x^2) tan(x-x^2) d/dx (x-x^2) #

The derivative of #(x-x^2) # is 1-2x, so:

#sec(x-x^2) tan(x-x^2) (1-2x) #