How do you differentiate y=(6+7x)/(14x+9)?

Oct 1, 2016

$\frac{7 \left(14 x + 9\right) - 14 \left(6 + 7 x\right)}{14 x + 9} ^ 2$

$\frac{98 x + 63 - 84 - 98 x}{14 x - 9} ^ 2$

-21/((14x-9)^2

Explanation:

This is just a basic quotient rule problem. The general formula for quotient rule is as follows:

For $f \frac{x}{g} \left(x\right)$:

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

Where $f \left(x\right)$ is the function in your numerator, and $g \left(x\right)$ is the function in your denominator.

Now, all we have to do is plug in everything.

$\implies \frac{\frac{d}{\mathrm{dx}} \left(6 + 7 x\right) \left(14 x + 9\right) - \frac{d}{\mathrm{dx}} \left(14 x + 9\right) \left(6 + 7 x\right)}{14 x + 9} ^ 2$

The derivatives you need to take here are quite simple, as both of your functions are linear. In any case, let's take them up:

$\frac{d}{\mathrm{dx}} \left(6 + 7 x\right) = 7$

$\frac{d}{\mathrm{dx}} \left(14 x + 9\right) = 14$

Now we just plug this back:

$\implies \frac{7 \left(14 x + 9\right) - 14 \left(6 + 7 x\right)}{14 x + 9} ^ 2$

Usually, it's best to leave this answer as is, as expanding doesn't get you anywhere. In this case, however, let's expand the numerator:

$\implies \frac{98 x + 63 - 84 - 98 x}{14 x - 9} ^ 2$

As you can see, the $98 x$'s cancel out. So, we're left with:

=>-21/((14x-9)^2