Hey there, Im working a problem finding the dy/dx for y=(3x-2)/(x^2+2x-3). I did the quotient rule and I'm stuck after simplifying the top.)

Question

zepdrix · Answer

$$\large y=\frac{3x-2}{x^2+2x-3}$$

$$\large y'=\frac{\color{royalblue}{(3x-2)'}(x^2+2x-3)-(3x-2)\color{royalblue}{(x^2+2x-3)'}}{(x^2+2x-3)^2}$$
$$\large y'=\frac{\color{royalblue}{(3)}(x^2+2x-3)-(3x-2)\color{royalblue}{(2x+2)}}{(x^2+2x-3)^2}$$

So you're having trouble simplifying it from here?