# Question #f2c8b

Jan 22, 2017

$y ' = \frac{3}{2 \left(3 x + 1\right)} + \frac{2}{5 - 2 x}$

#### Explanation:

First of all, separate the natural logarithms, using the rule $\ln \left(\frac{a}{b}\right) = \ln a - \ln b$.

$y = \ln \sqrt{3 x + 1} - \ln \left(5 - 2 x\right)$

$y = \ln {\left(3 x + 1\right)}^{\frac{1}{2}} - \ln \left(5 - 2 x\right)$

Apply the rule $\ln {a}^{n} = n \ln a$ to get rid of the exponent.

$y = \frac{1}{2} \ln \left(3 x + 1\right) - \ln \left(5 - 2 x\right)$

Now differentiate each natural logarithm separately using the chain rule.

For $\frac{d}{\mathrm{dx}} \frac{1}{2} \ln \left(3 x + 1\right)$

Let $y = \frac{1}{2} \ln u$ and $u = 3 x + 1$. Then $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 u}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 3$.

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

For $\frac{d}{\mathrm{dx}} \ln \left(5 - 2 x\right)$

Let $y = \ln u$ and $u = 5 - 2 x$. Then $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = - 2$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \cdot - 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{5 - 2 x}$

Now subtract them to find the derivative of the entire function.

$y ' = \frac{3}{2 \left(3 x + 1\right)} - \left(- \frac{2}{5 - 2 x}\right)$

$y ' = \frac{3}{2 \left(3 x + 1\right)} + \frac{2}{5 - 2 x}$

Hopefully this helps!

Jan 22, 2017

$y ' = \frac{6 x + 19}{- 12 {x}^{2} + 26 x + 10}$

#### Explanation:

We are dealing with two function compositions and a fraction, so we will require the use of the chain and quotient rules. The chain rule states that if $y$ is a function of $u$ and $u$ is a function of $x$, then

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

The quotient rule states that if $a , b$ two functions of $x$, then

$\frac{d \left(\frac{a}{b}\right)}{\mathrm{dx}} = \frac{a ' b - a b '}{b} ^ 2$

We know that $\left(\ln x\right) ' = \frac{1}{x}$, so if $u$ is a function of $x$, from the chain rule we have that $\left(\ln u\right) ' = \frac{1}{u} \cdot u '$:

$y ' = \frac{5 - 2 x}{\sqrt{3 x + 1}} \cdot \left[\frac{\sqrt{3 x + 1}}{5 - 2 x}\right] '$

Then, to calculate the derivative of the second term, use the quotient rule:

$\left[\frac{\sqrt{3 x + 1}}{5 - 2 x}\right] ' = \frac{\left(\sqrt{3 x + 1}\right) ' \left(5 - 2 x\right) - \left(\sqrt{3 x + 1}\right) \left(- 2\right)}{5 - 2 x} ^ 2$

Finally, we will have to find the derivative of $\sqrt{3 x + 1}$.

Using the chain rule, with $v = 3 x + 1$:

$\left(\sqrt{v}\right) ' = \frac{1}{2 \sqrt{v}} \cdot v ' = \frac{3}{2 \sqrt{3 x + 1}}$.

Substituting in the middle expression gives:

$\frac{\frac{3 \left(5 - 2 x\right)}{2 \sqrt{3 x + 1}} - \left(\sqrt{3 x + 1}\right) \left(- 2\right)}{5 - 2 x} ^ 2$

This looks quite messy. We can simplify by multiplying and dividing the second term in the numerator, by $2 \sqrt{3 x + 1}$, then transform the

fraction into a complex fraction of the form $\frac{\frac{a}{b}}{c}$:

$\frac{\frac{3 \left(5 - 2 x\right)}{2 \sqrt{3 x + 1}} - \frac{\left(3 x + 1\right) \left(- 4\right)}{2 \sqrt{3 x + 1}}}{5 - 2 x} ^ 2$

$= \frac{\frac{3 \left(5 - 2 x\right) - \left(3 x + 1\right) \left(- 4\right)}{2 \sqrt{3 x + 1}}}{5 - 2 x} ^ 2$

$= \frac{3 \left(5 - 2 x\right) - \left(3 x + 1\right) \left(- 4\right)}{\left(2 \sqrt{3 x + 1}\right) {\left(5 - 2 x\right)}^{2}}$

$= \frac{6 x + 19}{2 \left(\sqrt{3 x + 1}\right) {\left(5 - 2 x\right)}^{2}}$

Much better now. Finally, we can substitute this into the initial expression:

$y ' = \frac{5 - 2 x}{\sqrt{3 x + 1}} \cdot \frac{6 x + 19}{2 \left(\sqrt{3 x + 1}\right) {\left(5 - 2 x\right)}^{2}}$

$y ' = \frac{6 x + 19}{- 12 {x}^{2} + 26 x + 10}$

We could even factor the quadratic on the bottom to get:

$y ' = \frac{6 x + 19}{\left(x + \frac{1}{3}\right) \left(x - \frac{5}{2}\right)}$