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

substituting these results into f'(x)

To find the derivative of $f(x) = \frac{3 + x}{1 - 3x}$, you can use the quotient rule, which states that the derivative of a quotient of two functions is given by:

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

For the given function $f(x) = \frac{3 + x}{1 - 3x}$, let $u(x) = 3 + x$ and $v(x) = 1 - 3x$. Then, differentiate $u(x)$ and $v(x)$ to get $u'(x)$ and $v'(x)$, respectively:

$u'(x) = 1$ $v'(x) = -3$

Now apply the quotient rule:

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

Simplify the numerator and denominator:

$f'(x) = \frac{1 - 3x + 9 + 3x}{(1 - 3x)^2}$ $f'(x) = \frac{10}{(1 - 3x)^2}$

So, the derivative of $f(x) = \frac{3 + x}{1 - 3x}$ is $f'(x) = \frac{10}{(1 - 3x)^2}$.