Question

I got 8x/3cubroot of (1+8x). Is this right can someone explain the steps that they did to get the answer?

Answer 1

Answer:

$\frac{dy}{dx}[\sqrt[3]{1+8x} ] = \frac{8}{3\sqrt[3]{(1+8x)^2} }$

Explanation:

We need to use the chain rule, and you correctly identified it; however, you did not apply it correctly. Let's review it.

The chain rule is $f'(g(x)) * g'(x)$

To solve, we need $f'(x)$ and $g'(x)$. Let's find them.

$f'(x)= \frac{d}{du} [\sqrt[3]{u} ] = \frac{d}{du} [u^\frac{1}{3} ] = \frac{1}{3}u^\frac{-2}{3} = \frac{u^\frac{-2}{3} }{3} = \frac{1}{3u^\frac{2}{3} } \\g'(x)= \frac{d}{dx} [1+8x] = 8$

For both these derivatives, I used the power rule. For $f'(x)$, I just simplified it completely to make it look nicer. Now that we know everything, we need to plug it into the chain rule: $f'(g(x)) * g'(x)$

Take the $g(x)$ function and put it into $f'(x)$ and multiply all that by $g'(x)$.

$(\frac{1}{3(1+8x)^\frac{2}{3} } )(\frac{8}{1} ) = \frac{8}{3(1+8x)^\frac{2}{3} } = \frac{8}{3\sqrt[3]{(1+8x)^2} }$

So, your derivative is $\frac{8}{3\sqrt[3]{(1+8x)^2} }$