# Question 0b1d4

Apr 17, 2016

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

#### Explanation:

When differentiating logarithmically, first define the function:

y=3sqrt((x(x+1)(x-2))/((x^2+1)(2x+3))#

Now, take the natural logarithm of both sides.

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

One of logarithm's many abilities is to be able to be split up easily. Here, we can make use of the fact that terms being multiplied inside a logarithm can be split up into two separate logarithms being added, as follows:

$\ln \left(a b c\right) = \ln \left(a\right) + \ln \left(b\right) + \ln \left(c\right)$

This gives us

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

Before proceeding, we can also deal with the square root through another property of logarithms:

$\ln \left({a}^{b}\right) = b \cdot \ln \left(a\right)$

Thus, we have

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

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

Now, we can continue splitting up the logarithm. Recall that when there are terms being divided, such as $\left({x}^{2} + 1\right)$, it will be subtracted instead of added--an example being:

$\ln \left(\frac{a b}{c d}\right) = \ln \left(a\right) + \ln \left(b\right) - \ln \left(c\right) - \ln \left(d\right)$

Remembering that the $\frac{1}{2}$ will be distributed to each term, this yields:

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

Now, differentiate both sides of the equation. Recall that differentiation with the natural logarithm function takes the form:

$\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \frac{u '}{u}$

Thus, we obtain

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

Finally, to solve for $y '$, which is the derivative of the original function, multiply both sides of the equation by $y$. Recall that $y$ is the original function.

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