What is the derivative of sec(x-x^2)sec(xx2)?

2 Answers
Nov 9, 2016

d/dxsec(x-x^2) = (1-2x)sec(x-x^2)tan(x-x^2) ddxsec(xx2)=(12x)sec(xx2)tan(xx2)

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) 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)

Nov 9, 2016

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:


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)