Let $f(x)=(2x+1)^3$ and let $g$ be the inverse function of $f$. Given that $f(0)=1$, what is the value of $g'(1)$ ?

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For two functions to be inverses, this must be true:

$$f(g(x))=x$$

We can differentiate this equation to obtain the relationship between the derivatives of the function.

$$f'(g(x))\cdot g'(x) = 1$$

For this question, we are trying to find $g'(1)$.

$$g'(1)= \frac{1}{f'(g(1))}$$

Since the question mentions that $f(0)=1$, $g(1)=0$ must be true since $f$ and $g$ are inverses.

$$g'(1)=\frac{1}{f'(0)}$$

Find $f'(0)$:

$$f(x)=(2x+1)^3$$ $$f'(x)=6(2x+1)^2$$ $$f'(0)=6$$

Putting it all together:

$$g'(1)=\boxed{\frac{1}{6}}$$ We can swap $x$ and $y$ values to obtain the equation of the inverse function. $$y=(2x+1)^3$$ $$x=(2y+1)^3$$

Solve for $y$:

$$\sqrt[3]{x}=2y+1$$ $$y = \frac{\sqrt[3]{x}}{2}-\frac{1}{2}$$

The inverse function $g$ can be differentiated:

$$g(x)=\frac{1}{2}x^{\frac{1}{3}}-\frac{1}{2}$$ $$g'(x)=\frac{1}{6}x^{-\frac{2}{3}}$$ $$g'(1) = \frac{1}{6}(1)^{-\frac{2}{3}}$$ $$g'(1)= \boxed{\frac{1}{6}}$$