- #1

- 3,269

- 5

ok not real sure what the answer is but I did this (could be easier I am sure}

rewrite as

$y=(2x+1)^3$

exchange x and rename y to g

$x=(2g+1)^3$

Cube root each side

$\sqrt[3]{x}=2g+1$

isolate g

$g=\dfrac{\sqrt[3]{x}-1}{2}$

so

$\left(\dfrac{\sqrt[3]{x}-1}{2}\right)'

=\dfrac{1}{6x^{\dfrac{2}{3}}}$

apply $x=1$

$\dfrac{1}{6(1)^{\dfrac{2}{3}}}=\dfrac{1}{6}$

hopefully