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Question: Using the first principle of derivative, Evaluate \(\dfrac{{x + 1}}{{x - 1}}\)?...

Using the first principle of derivative, Evaluate x+1x−1\dfrac{{x + 1}}{{x - 1}}?

Explanation

Solution

We must use the first principle of differentiation to get the derivative of a function in order to answer the question. The first principle of differentiation aids in the evaluation of a function's derivative utilizing limits. Calculating a function's derivative using the first principle of differentiation can be time-consuming. To calculate the bounds and evaluate the value, we can use identities and tricks.

Complete answer:
We will use the fundamental principle of differentiation,
We will use the first principle of differentiation by using the formula lim⁡f′(x)=f(x+h)−f(x)h\lim f'(x) = \dfrac{{f(x + h) - f(x)}}{h}where h tends to 00.
When we multiply the fractions by their LCM, we obtain
f′(x)  =  limh→0[(x+h+1x+h−1)−(x+1x−1)h]f^{\prime } (x)\; = \;lim_{h \to 0}\left[ {\dfrac{{\left( {\dfrac{{x + h + 1}}{{x + h - 1}}} \right) - \left( {\dfrac{{x + 1}}{{x - 1}}} \right)}}{h}} \right]
We will first solve the numerator as a separate fractional part by taking LCM,
f′(x)  =  limh→0[((x+h+1)(x−1))−((x+1)(x+h−1))(x+h−1)(x−1)h]f^{\prime } (x)\; = \;lim_{h \to 0}\left[ {\dfrac{{\left( {\left( {x + h + 1} \right)\left( {x - 1} \right)} \right) - \left( {\left( {x + 1} \right)\left( {x + h - 1} \right)} \right)}}{{\left( {x + h - 1} \right)\left( {x - 1} \right)h}}} \right]
We will Open the brackets to multiply each part
f′(x)  =  limh→0[((x2−x+hx−h+x−1))−((x2+xh−x+x+h−1))(x+h−1)(x−1)h]f^{\prime } (x)\; = \;lim_{h \to 0}\left[ {\dfrac{{\left( {\left( {{x^2} - x + hx - h + x - 1} \right)} \right) - \left( {\left( {{x^2} + xh - x + x + h - 1} \right)} \right)}}{{\left( {x + h - 1} \right)\left( {x - 1} \right)h}}} \right]
f′(x)  =  limh→0[((x2−x+hx−h+x−1−x2−xh+x−x+h+1))(x+h−1)(x−1)h]f^{\prime } (x)\; = \;lim_{h \to 0}\left[ {\dfrac{{\left( {\left( {{x^2} - x + hx - h + x - 1 - {x^2} - xh + x - x + h + 1} \right)} \right)}}{{\left( {x + h - 1} \right)\left( {x - 1} \right)h}}} \right]
By further solving
f′(x)  =  limh→0[(−2h)(x+h−1)(x−1)h]f^{\prime } (x)\; = \;lim_{h \to 0}\left[ {\dfrac{{\left( { - 2h} \right)}}{{\left( {x + h - 1} \right)\left( {x - 1} \right)h}}} \right]
We first cancel h both from numerator and denominator and then put h=0, we get
f′(x)  =  [(−2)(x−1)(x−1)]f^{\prime } (x)\; = \;\left[ {\dfrac{{\left( { - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 1} \right)}}} \right]
So, the derivative of x+1x−1\dfrac{{x + 1}}{{x - 1}} is f′(x)  =  [(−2)(x−1)(x−1)]f^{\prime } (x)\; = \;\left[ {\dfrac{{\left( { - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 1} \right)}}} \right].

Note:
The L hospital rule of differentiation can also be used to find the derivative of a given function. The L Hospital rule is a way for evaluating indeterminate forms like 00\dfrac{0}{0}or ∞∞\dfrac{\infty }{\infty }. L'Hospital's rule is used in calculus to determine the limits of indeterminate forms for derivatives. The L Hospital Rule can be used several times. You can apply this rule indefinitely, and it will hold any indefinite form after each application. If the problem is out of the indeterminate forms, you can’t be able to apply L'Hospital Rule. We can answer the problem in ∞∞\dfrac{\infty }{\infty },00\dfrac{0}{0},∞0{\infty ^0},1∞{1^\infty }forms using the L Hospital rule.