# Simplifying a chain rule probelm

#### rcdavis28

I've got this chain rule problem and I want to make sure I understand the solution entirely.

7/(x3 +3x)1/3

I rewrite and apply the chain rule: 7 d/dx -1/3(x3 + 3x)-4/3 * (3x + 3)

My book has the answer as -7(x^2 + 1) / x^3 + 3x4/3

I just wanted to get an experts opinion on the simplification. Did they just use the 3 in the denominator of -1/3 to cancel out the 3's in the last part of the expression? (3x + 3). So, they factored out a 3 on the last expression and that 3 canceled out the 3 in the -1/3?

Thank you all in advance for any input!

#### romsek

MHF Helper
your lack of parens makes this hard to decipher

what's true is that

$\dfrac{d}{dx} \dfrac{7}{(x^3+3x)^{1/3}} = -\dfrac{7(x^2+1)}{(x^3+3x)^{4/3}}$

so let's look at why

\begin{align*} &\dfrac{d}{dx} \dfrac{7}{(x^3+3x)^{1/3}} =\\ \\ &\dfrac{d}{dx} 7(x^3+3x)^{-1/3} = \\ \\ &-\dfrac{7}{3}(x^3+3x)^{-4/3}\cdot (3x^2+3) = \\ \\ &-\dfrac{7(3x^2+3)}{3(x^3+3x)^{4/3}} = \\ \\ &-\dfrac{7(x^2+1)}{(x^3+3x)^{4/3}} \end{align*}

• rcdavis28

#### rcdavis28

Thanks. So I guess the 3 was factored out of the numerator and canceled with the 3 in the denominator.

MHF Helper
you guess right