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## Homework Statement

If [tex]y=\frac{1-2u}{1+u}[/tex] and [tex]u=\sqrt{x^2-7}[/tex], find [tex]\frac{dy}{dx}[/tex] at [tex]x=4[/tex]

## Homework Equations

quotient and chain rule

## The Attempt at a Solution

so if [tex]f=y[/tex] then [tex]y^{'}=f^{'}[/tex] then [tex]f^{'} = \left( \frac{f}{g}\right)^{'} = \frac{f^{'}g - fg^{'}}{g^2}[/tex]

and if [tex]f=1-2u[/tex] then [tex]f^{'} = -\left( x^2 -7\right)^\frac{-1}{2} 2x[/tex]

and if [tex]g=1+u[/tex] then [tex]g^{'} = x\left( x^2 -7\right)^\frac{-1}{2}[/tex]

and [tex]g^2 = x^2 + 2\sqrt{x^2 -7}-6[/tex]

and now with chain rule:

[tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\left(\frac{-2\left(1+u\right)-\left(1-2u\right)}{\left(1+u\right)^{2}}\right)\left(x\left(x^{2}-7\right)^{\frac{-1}{2}}\right)[/tex]

i think?

u in terms of x like this:

[tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\left(\frac{-2\left(1+\sqrt{x^2-7}\right)-\left(1-2\sqrt{x^2-7}\right)}{\left(1+\sqrt{x^2-7}\right)^{2}}\right)\left(x\left(x^{2}-7\right)^{\frac{-1}{2}}\right)[/tex]

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