Answer:
1/24
Explanation:
![f(x)=(-6)/(x)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/816sakhyts1oci40zoa4qkalrrhqyal7cc.png)
We want to find the derivative of f at x=12.
I'm assuming you want to see the formal definition of a derivative approach.
The definition is there:
![f'(x)=\lim_(h \rightarrow 0)(f(x+h)-f(x))/(h)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1m355swkw394lzt4ecl23inh99e1kbhlab.png)
So we need to find f(x+h) given f(x).
To do this all you have to is replace old input, x, with new input, (x+h).
Let's do that:
.
Let's go to the definition now:
![f'(x)=\lim_(h \rightarrow 0)(f(x+h)-f(x))/(h)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1m355swkw394lzt4ecl23inh99e1kbhlab.png)
![f'(x)=\lim_(h \rightarrow 0)((-6)/(x+h)-(-6)/(x))/(h)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/p60t11jg4tskrnvuaafb7ak53qheji64rc.png)
Multiply top and bottom by the least common multiple the denominators of the mini-fractions. That is, we are going to multiply top and bottom by x(x+h):
![f'(x)=\lim_(h \rightarrow 0)((-6)/(x+h)x(x+h)-(-6)/(x)x(x+h))/(hx(x+h))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/n6xn0d88s4myzu4y79cmbply1hb1a8bzaz.png)
Let's cancel the (x+h)'s in the first mini-fraction.
We will also cancel the (x)'s in the second-mini-fraction.
![f'(x)=\lim_(h \rightarrow 0)(-6x--6(x+h))/(hx(x+h))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zy1shhvnx1fsng2o4szxet895iz1rg11ne.png)
--=+ so I'm rewriting that part:
![f'(x)=\lim_(h \rightarrow 0)(-6x+6(x+h))/(hx(x+h))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vkfmq4x0cow7xir74qc2khy4jc7oawpncb.png)
Distribute (NOT ON BOTTOM!):
![f'(x)=\lim_(h \rightarrow 0)(-6x+6x+6h)/(hx(x+h))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rm5opxpqm20k8cirhcmdol28f363ip60gh.png)
Simplify the top (-6x+6x=0):
![f'(x)=\lim_(h \rightarrow 0)(6h)/(hx(x+h))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/w3gya87h7hq6db5i2k5niozl7xutd8k82b.png)
Simplify the fraction (h/h=1):
![f'(x)=\lim_(h \rightarrow 0)(6)/(x(x+h))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/86pkpgav4q8cqbk15nn44pkryigphsju2z.png)
Now you can plug in 0 for h because it doesn't give you 0/0:
![(6)/(x(x+0))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ee1gs8hgvczix4eo5tnc3ogi5tpbggwrnf.png)
![(6)/(x^2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mfitk2ikrx4kpqbea31sjwidzuzdpijjsw.png)
![f'(x)=(6)/(x^2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/sxgdmt7dylofjkcua1szyq0fc8d21qevnq.png)
We want to evaluated the derivative at x=12 so replace x with 12:
![f'(12)=(6)/(12^2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/om1mhnkqw8u5u7g78b01n7wxqxs64wj9b0.png)
![f'(12)=(6)/(144)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/u38n3xzb72h6i1bqgvj71xkji591lcs1nx.png)
Divide top and bottom by 6:
![f'(12)=(1)/(24)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/18lifq0frb662dxmfj71nyj2uriv2v6mv8.png)