Defferentiate f(x) = x-1/x+1 ? Thank you

Apr 20, 2018

$\frac{2}{{x}^{2} + 2 x + 1}$

Explanation:

quotient rule, for two different functions $u$ and $v :$

$\frac{\Delta y}{\Delta x} \left(\frac{u}{v}\right) = \frac{u ' v - v ' u}{{v}^{2}}$

if $u = x - 1$ and $v = x + 1 :$

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

$\frac{\Delta y}{\Delta x} \left(x + 1\right) = {x}^{0} + 0 = 1$

$\frac{\Delta y}{\Delta x} \left(x - 1\right) = {x}^{0} + 0 = 1$

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

$= \frac{x + 1 - x + 1}{{x}^{2} + 2 x + 1}$

$= \frac{2}{{x}^{2} + 2 x + 1}$

Apr 21, 2018

Through another method, we see $2 {\left(x + 1\right)}^{-} 2$ (this is no different, though, from the other answer provided)

Explanation:

Notice that the answer derived by LM, $\frac{2}{{x}^{2} + 2 x + 1}$, can be rewritten once you notice the denominator is factorable: in fact, it's the same as $\frac{2}{x + 1} ^ 2$, or $2 {\left(x + 1\right)}^{-} 2$.

This seems pretty coincidental, but we can see another way of differentiating the given function to see how to arrive at this answer more quickly.

Let's start by rewriting the numerator of the function in question:

$\frac{x - 1}{x + 1} = \frac{x + 1 - 2}{x + 1}$

We do this in order to split the function up and achieve some form of simplification in $\frac{x + 1}{x + 1}$:

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

Assuming you've learned the chain rule, this is relatively simple to differentiate. The derivative of $1$ is $0$, and the derivative of ${\left(x + 1\right)}^{-} 1$ is $- 1 {\left(x + 1\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right)$ by the chain rule (note that this is like the derivative of ${x}^{-} 1$).

We then see that:

$\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right) = 0 - 2 \left[- 1 {\left(x + 1\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right)\right]$

$= 2 {\left(x + 1\right)}^{-} 2 \cdot 1$

$= \frac{2}{x + 1} ^ 2$

As was found before!