Answer:
[tex]\frac{dy}{dx}[\sqrt[3]{1+8x} ] = \frac{8}{3\sqrt[3]{(1+8x)^2} }[/tex]
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 [tex]f'(g(x)) * g'(x)[/tex]
To solve, we need [tex]f'(x)[/tex] and [tex]g'(x)[/tex]. Let's find them.
[tex]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[/tex]
For both these derivatives, I used the power rule. For [tex]f'(x)[/tex], I just simplified it completely to make it look nicer. Now that we know everything, we need to plug it into the chain rule: [tex]f'(g(x)) * g'(x)[/tex]
Take the [tex]g(x)[/tex] function and put it into [tex]f'(x)[/tex] and multiply all that by [tex]g'(x)[/tex].
[tex](\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} }[/tex]
So, your derivative is [tex]\frac{8}{3\sqrt[3]{(1+8x)^2} }[/tex]
