Question

solve dy/dx of y=(x-1)diveide(x+1)
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Answer 1

I assume you're reffering to 5.b.

If

$y = \left(\dfrac{x-1}{x+1}\right)^{1/3}$

then we can rewrite this as

$y^3 = \dfrac{x-1}{x+1} \\\\ \ln(y^3) = \ln\left(\dfrac{x-1}{x+1}\right) \\\\ 3\ln(y) = \ln(x-1) - \ln(x+1)$

Now differentiate both sides implicitly:

$\dfrac{\mathrm d}{\mathrm dx}\left[3\ln(y(x))\right] = \dfrac{\mathrm d}{\mathrm dx}\left[\ln(x-1)-\ln(x+1)\right] \\\\ \dfrac3{y(x)}\dfrac{\mathrm dy}{\mathrm dx} = \dfrac1{x-1} - \dfrac1{x+1} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{y(x)}3\left(\dfrac1{x-1}-\dfrac1{x+1}\right) \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac13 \left(\dfrac{x-1}{x+1}\right)^{1/3} \left(\dfrac1{x-1}-\dfrac1{x+1}\right) \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \boxed{\dfrac23 \dfrac1{x^2-1}\left(\dfrac{x-1}{x+1}\right)^{1/3}}$