\displaystyle f(x)=\frac{aplusbx}{pplusqx}.
Compute [f(x+h)]-[f(x)].\displaystyle
\frac{Nh}{(pplusqxplusqh)(pplusqx)}
,N is an integer.
(You only enter N into the Answer box.)h, and then take \displaystyle\lim_{h\to0}.
This gives f'(x).
\quad\quadpbminusaqh
\displaystyle\frac{\phantom{X}}
{\phantom{(pplusqxplusqh)(pplusqx)}}
\,(pplusqxplusqh)(pplusqx)
Let X=[f(x+h)]-[f(x)].
Then \displaystyle X=
\left[\frac{aplusbxplusbh}
{pplusqxplusqh}\right]
-\left[\frac{aplusbx}{pplusqx}\right]
\displaystyle=
\frac{(aplusbxplusbh)(pplusqx)
-(aplusbx)(pplusqxplusqh)}
{(pplusqxplusqh)(pplusqx)}.
Let A=(aplusbxplusbh)(pplusqx)
and B=(aplusbx)(pplusqxplusqh)
and C=(pplusqxplusqh)(pplusqx).
Then \displaystyle X=\frac{A-B}{C}.
A=(aplusbx)(pplusqx)+
(bh)(pplusqx) and
B=(aplusbx)(pplusqx)+
(aplusbx)(qh).
Next, we compute A-B.
The terms (aplusbx)(pplusqx)
in A and B cancel,
and we get
A-B=
(bh)(pplusqx)-
(aplusbx)(qh).
Expanding,
A-B=
(bh)(p)+(bh)(qx)-
(a)(qh)-(bx)(qh).
The terms (bh)(qx) and -(bx)(qh)
cancel, and we get
A-B=(bh)(p)-(a)(qh).
Then
A-B=pbhminusaqh=pbminusaqh.
Then \displaystyle X=\frac{A-B}{C}=
\frac{pbminusaqh}
{(pplusqxplusqh)(pplusqx)}.
Now enter pbminusaq into the Answer box, to indicate that the answer is
\displaystyle\frac{pbminusaqh}
{(pplusqxplusqh)(pplusqx)}.
NOT FOR CREDIT, BUT FOR EXTRA LEARNING:
Recall that X=[f(x+h)]-[f(x)]. In the next hint,
we compute \displaystyle\frac{X}{h} (on h\ne0),
and then \displaystyle f'(x)=\lim_{h\to0}\frac{X}{h}.
For all h\ne0,
\displaystyle\frac{X}{h}=\frac{pbminusaq}
{(pplusqxplusqh)(pplusqx)}.
\displaystyle f'(x)=\lim_{h\to0}\frac{X}{h}=
\frac{pbminusaq}{(pplusqx)^2}.