# What is the first and second derivative of (10x*(x-1)^4)/((x-2)^3*(x+1)^2) ?

Oct 31, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\frac{1}{x} + \frac{4}{x - 1} - \frac{3}{x - 2} - \frac{2}{x + 1}\right] \cdot \frac{10 x \cdot {\left(x - 1\right)}^{4}}{{\left(x - 2\right)}^{3} \cdot {\left(x + 1\right)}^{2}}$

#### Explanation:

As there are many power in this question and it is hard to differentiate this form, we can use the logarithmic differentiation to determine the derivative.

First, we know that

1. $\ln \left(a \cdot b\right) = \ln a + \ln b$
2. $\ln \left(\frac{a}{b}\right) = \ln a - \ln b$
3. $\ln \left({a}^{n}\right) = n \ln a$

We can let $y = \frac{10 x \cdot {\left(x - 1\right)}^{4}}{{\left(x - 2\right)}^{3} \cdot {\left(x + 1\right)}^{2}}$
Then, $\ln y = \ln \left(10 x\right) + \ln \left[{\left(x - 1\right)}^{4}\right] - \left\{\left[\ln {\left(x - 2\right)}^{3}\right] + \ln {\left(x + 1\right)}^{2}\right\}$

$\ln y = \ln \left(10 x\right) + 4 \ln \left(x - 1\right) - \left[3 \ln \left(x - 2\right) + 2 \ln \left(x + 1\right)\right]$

$\ln y = \ln \left(10 x\right) + 4 \ln \left(x - 1\right) - 3 \ln \left(x - 2\right) - 2 \ln \left(x + 1\right)$

It is more easy for us to differentiate this form now.
And we also know that $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x} \cdot \frac{d}{\mathrm{dx}} \left(x\right) = \frac{1}{x}$

So, we can differentiate both sides to get the answer:
$\frac{d}{\mathrm{dx}} \left(\ln y\right) = \frac{d}{\mathrm{dx}} \left[\ln \left(10 x\right) + 4 \ln \left(x - 1\right) - 3 \ln \left(x - 2\right) - 2 \ln \left(x + 1\right)\right]$

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{10 x} \cdot \frac{d}{\mathrm{dx}} \left(10 x\right) + 4 \cdot \frac{1}{x - 1} \cdot \frac{d}{\mathrm{dx}} \left(x - 1\right) - 3 \cdot \frac{1}{x - 2} \cdot \frac{d}{\mathrm{dx}} \left(x - 2\right) - 2 \cdot \frac{1}{x + 1} \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right)$

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{10 x} \cdot 10 + \frac{4}{x - 1} \cdot 1 - \frac{3}{x - 2} \cdot 1 - \frac{2}{x + 1} \cdot 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\frac{1}{x} + \frac{4}{x - 1} - \frac{3}{x - 2} - \frac{2}{x + 1}\right] \cdot y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\frac{1}{x} + \frac{4}{x - 1} - \frac{3}{x - 2} - \frac{2}{x + 1}\right] \cdot \frac{10 x \cdot {\left(x - 1\right)}^{4}}{{\left(x - 2\right)}^{3} \cdot {\left(x + 1\right)}^{2}}$

This is the answer for this question :)