Question

y=(3x-7)/(2x+4).
What is dy/dx?

Answer 1

Remember the derivative of a product:
\[(f.g)'(x) = f'(x)g(x)+f(x)g'(x)\]

Answer 2

(f/g)'(x) = [f'(x)g(x)-f(x)g'(x)]/[g(x)^2] The quotient rule.
I have worked it through a bunch of times but I am not getting the right answer

Answer 3

\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=3x-7,f'(x)=3,g(x)=2x+4,g'(x)=2\]

Answer 4

oh... It was a quotient...
I'll help you.
So we have:
\[y= \frac{ (3x-7) }{ (2x+4) }\]
so then y' must be:
\[y'=\frac{ (3)(2x+4)-(3x-7)(2) }{ (2x+4)^{2} }\]
All you have to do from here on is simplify that equation.

Answer 5

I get 26/[(2x+4)^2] but webassign is telling me that its wrong

Answer 6

True, see we have (2x+4) ^2 on the denominator, and a (2x+4) multiplying in the numerator, so we can simplify them, ending up with:
\[y'= \frac{ 3-(3x-7)(2) }{ (2x+4) }\]

Answer 7

I don't think you can do that

Answer 8

answer is 13/[2(x+2)^2] but I still don't know how to get this answer