Question

find the derivative of the function y = √(-8 + 9x).
\frac{dy}{dx}=square

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt{-8 + 9x}$ as $y=(-8 + 9x)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=-8 + 9x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$. Second, find $\frac{du}{dx}$: $\frac{du}{dx}=9$.

Step3: Substitute and simplify

Substitute $u=-8 + 9x$ back into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$: $\frac{dy}{dx}=\frac{1}{2}(-8 + 9x)^{-\frac{1}{2}}\cdot9=\frac{9}{2\sqrt{-8 + 9x}}$.

Answer:

$\frac{9}{2\sqrt{-8 + 9x}}$