# How do you find the derivative of y = ln[(x+3)/x^2)]?

Dec 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x + 6}{x \left(x + 3\right)}$

#### Explanation:

We let $y = \ln u$ and $u = \frac{x + 3}{x} ^ 2$

Then $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1 \left({x}^{2}\right) - 2 x \left(x + 3\right)}{{x}^{2}} ^ 2 = \frac{{x}^{2} - 2 {x}^{2} - 6 x}{x} ^ 4 = \frac{- x \left(x + 6\right)}{x} ^ 4 = - \frac{x + 6}{x} ^ 3$

The chain rule states that color(red)(dy/dx= dy/(du) xx (du)/dx.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \times - \frac{x + 6}{x} ^ 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \frac{x + 6}{x} ^ 3}{\frac{x + 3}{x} ^ 2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x + 6}{x} ^ 3 \times {x}^{2} / \left(x + 3\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x + 6}{x \left(x + 3\right)}$

Hopefully this helps!

Dec 2, 2016

We can also rewrite this the logarithm rules $\log \left(\frac{a}{b}\right) = \log \left(a\right) - \log \left(b\right)$ and $\log \left({c}^{d}\right) = \mathrm{dl} o g \left(c\right)$.

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

$\textcolor{w h i t e}{y = \ln \left(\frac{x + 3}{x} ^ 2\right)} = \ln \left(x + 3\right) - 2 \ln \left(x\right)$

We should know that $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$. Applying the chain rule to this shows us that $\frac{d}{\mathrm{dx}} \ln \left(u\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x + 3} \cdot \frac{d}{\mathrm{dx}} \left(x + 3\right) - 2 \left(\frac{1}{x}\right)$

Since $\frac{d}{\mathrm{dx}} \left(x + 3\right) = 1$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x + 3} - \frac{2}{x}$

Finding a common denominator:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - 2 \left(x + 3\right)}{x \left(x + 3\right)}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{- x - 6}{x \left(x + 3\right)}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = - \frac{x + 6}{x \left(x + 3\right)}$