Question

Use both the definition of derivative and the alternate definition of derivative to find the derivativef(x)=1/1-x

Answer 1

Using both the definition of a derivative and rules of differentiation, we find that the derivative of the function f(x) = 1/(1-x) is f'(x) = 1/(1-x)^2 or f'(x) = -(1-x)^{-2}.

Step-by-step explanation:

To find the derivative of the function f(x) = 1/(1-x), we can use both the definition of a derivative and the alternate definition of a derivative. The definition of a derivative, also known as the limit of the difference quotient, is:

f'(x) = lim_{h\to 0} (f(x+h) - f(x))/h

Using this definition, we substitute f(x):

f'(x) = lim_{h\to 0} (\frac{1}{1-(x+h)} - \frac{1}{1-x})/h

We then find a common denominator and simplify:

f'(x) = lim_{h\to 0} (\frac{1-x - (1-(x+h))}{(1-x)(1-(x+h))})/h

f'(x) = lim_{h\to 0} (\frac{h}{(1-x)(1-x-h)})/h

After canceling h, we simplify further to obtain:

f'(x) = lim_{h\to 0} \frac{1}{(1-x)^2} = \frac{1}{(1-x)^2}

The alternate definition of a derivative involves using rules of differentiation applied directly, in this case, the quotient rule or recognizing that f(x) can be rewritten with a negative exponent:

f(x) = (1-x)^{-1}

Then we can apply the power rule:

f'(x) = -1*(1-x)^{-1 - 1}

f'(x) = -1*(1-x)^{-2}

Both methods give us the same result for the derivative of f(x):

f'(x) = \frac{1}{(1-x)^2} or f'(x) = -(1-x)^{-2}

Answer 2

Given the function

`f(x)=(1)/(1-x)`

The derivative of the function will be as follows:

`\begin{gathered} f^(\prime)(x)=((1-x)\cdot0-1\cdot(0-1))/((1-x)^2) \\ \\ f^(\prime)(x)=(1)/((1-x)^2) \end{gathered}`

Another alternate method:

the limit definition of the derivative will be as follows:

We will find f(x+h)

so,

`f(x+h)=(1)/((1-x-h))`

We will the difference between f(x+h) and f(x)

`\begin{gathered} f(x+h)-f(x)=(1)/(1-x-h)-(1)/(1-x) \\ f(x+h)-f(x)=((1-x)-(1-x-h))/((1-x-h)(1-x)) \\ f(x+h)-f(x)=(1-x-1+x+h)/((1-x-h)(1-x)) \\ f(x+h)-f(x)=(h)/((1-x-h)(1-x)) \end{gathered}`

Then divide the last result by h:

`\begin{gathered} (f(x+h)-f(x))/(h)=(h)/((1-x-h)(1-x))\cdot(1)/(h) \\ \\ (f(x+h)-f(x))/(h)=(1)/((1-x-h)(1-x)) \end{gathered}`

And finally, we will find the limit when h→0

so,

`\lim _(h\rightarrow0)(f(x+h)-f(x))/(h)=\lim _(h\rightarrow0)(1)/((1-x-h)(1-x))=(1)/((1-x)(1-x))=(1)/((1-x)^2)`

the alternative definition to find the derivative:

`\lim _(x\rightarrow a)(f(x)-f(a))/(x-a)`

We will find the value of f(a), this means we will substitute with (x = a) into the given function

so,

`f(a)=(1)/(1-a)`

Now, find the difference between f(x) and f(a)

`\begin{gathered} f(x)-f(a)=(1)/(1-x)-(1)/(1-a) \\ \\ f(x)-f(a)=((1-a)-(1-x))/((1-x)(1-a)) \\ \\ f(x)-f(a)=(1-a-1+x)/((1-x)(1-a))_{} \\ \\ f(x)-f(a)=(x-a)/((1-x)(1-a)) \end{gathered}`

divide the last result by (x-a)

so,

`\begin{gathered} (f(x)-f(a))/(x-a)=((x-a))/((x-a)(1-x)(1-a)) \\ \\ (f(x)-f(a))/(x-a)=(1)/((1-x)(1-a)) \end{gathered}`

Finally, find the limit when x→a

So,

`f^(\prime)(a)=\lim _(x\rightarrow a)(1)/((1-x)(1-a))=(1)/((1-a)(1-a))=(1)/((1-a)^2)`