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

Nov 9, 2016

$\frac{d}{\mathrm{dx}} \sec \left(x - {x}^{2}\right) = \left(1 - 2 x\right) \sec \left(x - {x}^{2}\right) \tan \left(x - {x}^{2}\right)$

#### Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If $y = f \left(x\right)$ then $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

I was taught to remember that the differential can be treated like a fraction and that the "$\mathrm{dx}$'s" of a common variable will "cancel" (It is important to realise that $\frac{\mathrm{dy}}{\mathrm{dx}}$ isn't a fraction but an operator that acts on a function, there is no such thing as "$\mathrm{dx}$" or "$\mathrm{dy}$" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$ etc, or $\left(\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\textcolor{red}{\cancel{\mathrm{dv}}}} \frac{\textcolor{red}{\cancel{\mathrm{dv}}}}{\textcolor{b l u e}{\cancel{\mathrm{du}}}} \frac{\textcolor{b l u e}{\cancel{\mathrm{du}}}}{\mathrm{dx}}\right)$

So with $y = \sec \left(x - {x}^{2}\right)$, Then:

$\left\{\begin{matrix}\text{Let "u= & => & (du)/dx=1-2x \\ "Then } y = \sec u & \implies & \frac{\mathrm{dy}}{\mathrm{du}} = \sec u \tan u\end{matrix}\right.$

(NB you should know that $\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x}$ ; If you don't then learn it!)

Using $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dy}}{\mathrm{du}}\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$ we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec u \tan u \left(1 - 2 x\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 - 2 x\right) \sec \left(x - {x}^{2}\right) \tan \left(x - {x}^{2}\right)$

Nov 9, 2016

$\sec \left(x - {x}^{2}\right) \tan \left(x - {x}^{2}\right) \left(1 - 2 x\right)$

#### Explanation:

you start of with:
$\sec \left(x - {x}^{2}\right)$

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)=$\frac{1}{\cos} \left(x\right)$

since we have $\frac{1}{\cos} \left(x\right)$, we will use the quotient rule, which states:

$\frac{g}{h}$ = $\frac{g ' h - h ' g}{g} ^ 2$

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

$\frac{\cos \left(x\right) \left(0\right) - \left(1\right) - \sin \left(x\right)}{\cos} {\left(x\right)}^{2}$

$\sin \frac{x}{\cos} {\left(x\right)}^{2}$

$\frac{1}{\cos} \left(x\right) \cdot \sin \frac{x}{\cos} \left(x\right)$

$\frac{1}{\cos} \left(x\right)$ = sec(x) and $\sin \frac{x}{\cos} \left(x\right)$ = tan(x)

Continuing on with the previous discussion
sec(x) is sex(x)tan(x), so:
$\sec \left(x - {x}^{2}\right) \tan \left(x - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left(x - {x}^{2}\right)$

The derivative of $\left(x - {x}^{2}\right)$ is 1-2x, so:

$\sec \left(x - {x}^{2}\right) \tan \left(x - {x}^{2}\right) \left(1 - 2 x\right)$