# How do you differentiate f(x)=sqrt(ln((x+3)^4 / (x+x^2))) using the chain rule?

Jan 7, 2016

$f ' \left(x\right) = \frac{1}{2 \sqrt{4 \ln \left(x + 3\right) - \ln \left(x + {x}^{2}\right)}} \cdot \left(\frac{4}{x + 3} - \frac{1 + 2 x}{x + {x}^{2}}\right)$.

#### Explanation:

Firstly, I'd try to simplify $f \left(x\right)$ in the following way, using log laws.

$f \left(x\right) = \sqrt{\ln \left({\left(x + 3\right)}^{4}\right) - \ln \left(x + {x}^{2}\right)}$

$f \left(x\right) = \sqrt{4 \ln \left(x + 3\right) - \ln \left(x + {x}^{2}\right)}$

$f \left(x\right) = {\left(4 \ln \left(x + 3\right) - \ln \left(x + {x}^{2}\right)\right)}^{\frac{1}{2}}$

Now let's try to differentiate, the derivative of the outside with respect to the inside, which is the square root function is denoted in red, the derivative of the radicand i.e., logs is in teal. Of course, we can think of doing the chain rule inside the log functions, which I've shown in purple.

$f ' \left(x\right) = \textcolor{red}{\left(\frac{1}{2}\right) \cdot {\left(4 \ln \left(x + 3\right) - \ln \left(x + {x}^{2}\right)\right)}^{- \frac{1}{2}}} \cdot \textcolor{t e a l}{\left(\frac{\textcolor{p u r p \le}{4}}{x + 3} - \frac{\textcolor{p u r p \le}{\left(1 + 2 x\right)}}{x + {x}^{2}}\right)}$.

Writing the index in radical form, we have:

$f ' \left(x\right) = \frac{1}{2 \sqrt{4 \ln \left(x + 3\right) - \ln \left(x + {x}^{2}\right)}} \cdot \left(\frac{4}{x + 3} - \frac{1 + 2 x}{x + {x}^{2}}\right)$, or

$f ' \left(x\right) = \frac{1}{2 \sqrt{\ln \left({\left(x + 3\right)}^{4} / \left(x + {x}^{2}\right)\right)}} \cdot \left(\frac{4}{x + 3} - \frac{1 + 2 x}{x + {x}^{2}}\right)$. Whichever looks more aesthetically pleasing.