Step 1
Before finding the derivative, it is helpful to rewrite the function with an expression raised to a power.
f(t) = \sqrt{9t^2 - t}
= \left(9t^2 - t\right)^{1/2}
Step 2
We wish to find $f'(t) = \frac{d}{dt}\left[(9t^2 - t)^{1/2}\right]$.
Recall the general power rule, which states that if the function $f$ is differentiable and $h(x) = \left[f(x)\right]^n$, where $n$ is a real number, then the following is true.
$h'(x) = \frac{d}{dx}\left[f(x)\right]^n = n\left[f(x)\right]^{n-1}f'(x)$
For $\frac{d}{dt}\left[(9t^2 - t)^{1/2}\right]$ we have a differentiable function being raised to a power. Let $g(t) = 9t^2 - t$ and $n = \frac{1}{2}$. Therefore, we have the following.
$\frac{d}{dt}\left[\left(g(t)\right)^{1/2}\right] = \frac{d}{dt}\left[(9t^2 - t)^{1/2}\right]$
Applying the general power rule to the above gives us the following result.
$\frac{d}{dt}\left[\left(g(t)\right)^n\right] = n\left[g(t)\right]^{n-1}g'(t)$
$\frac{d}{dt}\left[(9t^2 - t)^{1/2}\right] = \frac{1}{2}(9t^2 - t)^{1/2 - 1}\left(9(2t) - 1\right)$
To conclude, state the derivative $f'(t)$.
f'(t) = \frac{d}{dt}\left[(9t^2 - t)^{1/2}\right]
