How do you find the derivative of f(x)=(3+x)/(1-3x)?

Mar 3, 2016

$f = 3 + x , g = 1 - 3 x , f ' = 1 , g ' = - 3$
f'(x)=((gf'-fg')/g^2) =( (1-3x)-(-3(3+x)))/(1-3x)^2->(1-3x+9+3x)/(1-3x)^2=10/((1-3x)^2

Explanation:

Separate the top and the bottom into f and g then find the derivatives of each of them and then put it into the quotient rule $\frac{g f ' + f g '}{g} ^ 2$

Mar 3, 2016

$f ' \left(x\right) = - \frac{10}{1 - 3 x} ^ 2$

Explanation:

differentiate using the$\textcolor{b l u e}{\text{ Quotient rule }}$

If f(x)$= g \frac{x}{h \left(x\right)} \text{ then } f ' \left(x\right) = \frac{g \left(x\right) . h ' \left(x\right) - h \left(x\right) . g ' \left(x\right)}{h \left(x\right)} ^ 2$

here : g(x) = 3+x$\text{ and } g ' \left(x\right) = 1$

and $h \left(x\right) = 1 - 3 x \text{ and } h ' \left(x\right) = - 3$

substituting these results into f'(x)

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

$= \frac{- 9 - 3 x - 1 + 3 x}{1 - 3 x} ^ 2 = - \frac{10}{1 - 3 x} ^ 2$