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Math Help - differentiation - chain rule

  1. #1
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    differentiation - chain rule

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

    here's my working:
    du/dx = -\frac 4{(1+2x)^2}
    dy/du = \frac 1 2 sec^{\frac 1 2 } u tan u
    \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.
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    bk2 p53 re 17 q33b
    question:
    if u = \frac {1-2x}{1+2x} , find du/dx
    suppose that  y^2 = secu , where u = \frac {1-2x}{1+2x} , express dy/dx in terms of u

    here's my working:
    du/dx = -\frac 4{(1+2x)^2}
    dy/du = \frac 1 2 sec^{\frac 1 2 } u tan u
    \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
    u = \frac{1 - 2x}{1 + 2x} then

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

    etc.

    -Dan
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