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.