1. differentiation - chain rule

bk2 p53 re 17 q33b
question:
if $\displaystyle u = \frac {1-2x}{1+2x}$ , find du/dx
suppose that$\displaystyle y^2 = secu$ , where $\displaystyle u = \frac {1-2x}{1+2x}$ , express dy/dx in terms of u

here's my working:
$\displaystyle du/dx = -\frac 4{(1+2x)^2}$
$\displaystyle dy/du = \frac 1 2 sec^{\frac 1 2 } u tan u$
$\displaystyle \frac {dy}{dx} = \frac {dy}{du } \cdot \frac {du }{dx} = -\frac 4{(1+2x)^2} \frac 1 2 sec^{\frac 1 2 } u tan u$
don't know how to change the x in terms of u. thanks.

2. Originally Posted by afeasfaerw23231233
bk2 p53 re 17 q33b
question:
if $\displaystyle u = \frac {1-2x}{1+2x}$ , find du/dx
suppose that$\displaystyle y^2 = secu$ , where $\displaystyle u = \frac {1-2x}{1+2x}$ , express dy/dx in terms of u

here's my working:
$\displaystyle du/dx = -\frac 4{(1+2x)^2}$
$\displaystyle dy/du = \frac 1 2 sec^{\frac 1 2 } u tan u$
$\displaystyle \frac {dy}{dx} = \frac {dy}{du } \cdot \frac {du }{dx} = -\frac 4{(1+2x)^2} \frac 1 2 sec^{\frac 1 2 } u tan u$
don't know how to change the x in terms of u. thanks.
Well if
$\displaystyle u = \frac{1 - 2x}{1 + 2x}$ then

$\displaystyle \sqrt{sec(u)} = \sqrt{sec \left ( \frac{1 - 2x}{1 + 2x} \right ) }$

etc.

-Dan