Question 1187972
<font face="Times New Roman" size="+2">


Let *[tex \Large g(x)\ =\ x] and *[tex \Large h(x)\ =\ 4x\,-\,6], then *[tex \Large f(x)\ =\ \frac{g(x)}{h(x)}]


Quotient Rule:  If *[tex \Large f(x)\ =\ \frac{g(x)}{h(x)}], then *[tex \Large f'(x)\ =\ \frac{g'(x)h(x)\,-\,h'(x)g(x)}{\[ h(x) \]^2}]


Therefore, since *[tex \Large f(t)\ =\ \frac{t}{4t\,-\,6}]: 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(t)\ =\ \frac{g'(t)h(t)\,-\,h'(t)g(t)}{\[ h(t) \]^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(t)\ =\ \frac{4t\,-\,6\ -\ 4t}{(4t\,-\,6)^2}]


You can simplify it for yourself.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>