Respuesta :
Given the function
[tex]f(x)=\frac{1}{1-x}[/tex]The derivative of the function will be as follows:
[tex]\begin{gathered} f^{\prime}(x)=\frac{(1-x)\cdot0-1\cdot(0-1)}{(1-x)^2} \\ \\ f^{\prime}(x)=\frac{1}{(1-x)^2} \end{gathered}[/tex]Another alternate method:
the limit definition of the derivative will be as follows:
We will find f(x+h)
so,
[tex]f(x+h)=\frac{1}{(1-x-h)}[/tex]We will the difference between f(x+h) and f(x)
[tex]\begin{gathered} f(x+h)-f(x)=\frac{1}{1-x-h}-\frac{1}{1-x} \\ f(x+h)-f(x)=\frac{(1-x)-(1-x-h)}{(1-x-h)(1-x)} \\ f(x+h)-f(x)=\frac{1-x-1+x+h}{(1-x-h)(1-x)} \\ f(x+h)-f(x)=\frac{h}{(1-x-h)(1-x)} \end{gathered}[/tex]Then divide the last result by h:
[tex]\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{h}{(1-x-h)(1-x)}\cdot\frac{1}{h} \\ \\ \frac{f(x+h)-f(x)}{h}=\frac{1}{(1-x-h)(1-x)} \end{gathered}[/tex]And finally, we will find the limit when h→0
so,
[tex]\lim _{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=\lim _{h\rightarrow0}\frac{1}{(1-x-h)(1-x)}=\frac{1}{(1-x)(1-x)}=\frac{1}{(1-x)^2}[/tex]the alternative definition to find the derivative:
[tex]\lim _{x\rightarrow a}\frac{f(x)-f(a)}{x-a}[/tex]We will find the value of f(a), this means we will substitute with (x = a) into the given function
so,
[tex]f(a)=\frac{1}{1-a}[/tex]Now, find the difference between f(x) and f(a)
[tex]\begin{gathered} f(x)-f(a)=\frac{1}{1-x}-\frac{1}{1-a} \\ \\ f(x)-f(a)=\frac{(1-a)-(1-x)}{(1-x)(1-a)} \\ \\ f(x)-f(a)=\frac{1-a-1+x}{(1-x)(1-a)}_{} \\ \\ f(x)-f(a)=\frac{x-a}{(1-x)(1-a)} \end{gathered}[/tex]divide the last result by (x-a)
so,
[tex]\begin{gathered} \frac{f(x)-f(a)}{x-a}=\frac{(x-a)}{(x-a)(1-x)(1-a)} \\ \\ \frac{f(x)-f(a)}{x-a}=\frac{1}{(1-x)(1-a)} \end{gathered}[/tex]Finally, find the limit when x→a
So,
[tex]f^{\prime}(a)=\lim _{x\rightarrow a}\frac{1}{(1-x)(1-a)}=\frac{1}{(1-a)(1-a)}=\frac{1}{(1-a)^2}[/tex]