# Find the differential dy of the given function. (Use "dx" for dx.) y = \frac{x + 3}{3x - 4}

## Question:

Find the differential {eq}dy {/eq} of the given function. (Use "{eq}dx {/eq}" for {eq}dx {/eq}.)

{eq}\displaystyle\; y = \frac{x + 3}{3x - 4} {/eq}

## Quotient Rule:

The differential of {eq}y=f(x) {/eq} is {eq}\mathrm{d}y = f'(x) \ \mathrm{d}x {/eq} and if {eq}f(x) {/eq} is a quotient, the quotient rule must be employed to get the differential of {eq}y = f(x) {/eq}.

The formula for the quotient rule is indicated below:

{eq}\displaystyle \left(\frac{f}{g}\right)= \frac{gf'-fg'}{g^2} {/eq}

We will calculate the differential of {eq}\displaystyle\; y = \frac{x + 3}{3x - 4} {/eq} by dint of the quotient rule (note that we can simply differentiate {eq}y {/eq} using quotient rule and then merely rearrange it to find the differential):

{eq}\begin{align*} \displaystyle\; y& = \frac{x + 3}{3x - 4}\\ \mathrm{d}y& =\frac{(3x - 4)(x+3)' - (x+3)(3x-4)'}{(3x-4)^2} \ \mathrm{d}x && \left[\mathrm{ Quotient \ Rule }\right]\\ \mathrm{d}y& =\frac{(3x - 4)(1) - (x+3)(3)}{(3x-4)^2} \ \mathrm{d}x\\ \mathrm{d}y& =-\frac{13}{(3x-4)^2} \ \mathrm{d}x\\ \end{align*} {/eq}

Thus, the differential is {eq}\displaystyle \mathrm{d}y =-\frac{13}{(3x-4)^2} \ \mathrm{d}x {/eq}. 