Question

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

Answer 1

`d/dxsec(x-x^2) = (1-2x)sec(x-x^2)tan(x-x^2)`

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 = 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)`

Answer 2

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

Explanation:

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

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

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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)

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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)`