Question

Find the f'(x) of (3y-2x)-(8y-3x)

Answer 1

\[\frac{\partial \text{}}{\partial x}((3 y-2 x)-(8 y-3 x))=1\]

Answer 2

o sorry it it is suppose to be (3y-2x)/(8y-3x)

Answer 3

\[\text{Use the quotient rule, } \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v \frac{du}{dx}-u \frac{dv}{dx}}{v^2} ,\]
\[\text{where } u=3 y-2 x \text{ and } v=8 y-3 x\]
\[=\frac{(8 y-3 x) \left(\frac{d}{dx}(3 y-2 x)\right)-(3 y-2 x) \left(\frac{d}{dx}(8 y-3 x)\right)}{(8 y-3 x)^2}\]
\[\frac{-2 (8 y-3 x)-(8*0-3) (3 y-2 x)}{(8 y-3 x)^2}\]
\[-\frac{7 y}{(3 x-8 y)^2}\]

Answer 4

When you take the d/dx of (3y- 2x) why is it just -2 and not (3-2)