Question

How do you find the derivative of $y=\dfrac{x+1}{x-1}$?

Answer 1

Hint: To differentiate the function $y=\dfrac{x+1}{x-1}$ use quotient rule. Quotient rule states that if there are two functions say \[f\left( x \right)\] and \[g\left( x \right)\] then the derivative of \[\dfrac{f\left( x \right)}{g\left( x \right)}\] will be\[\dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{g\left( x \right)\dfrac{df\left( x \right)}{dx}-f\left( x \right)\dfrac{dg\left( x \right)}{dx}}{{{\left( g\left( x \right) \right)}^{2}}}\]. Furthermore differentiate the functions \[f\left( x \right)\] and \[g\left( x \right)\] separately. Hence if the solution is reducible then reduced it to more simplified form.

Complete step by step solution:
We have the given function $y=\dfrac{x+1}{x-1}.........\left( i \right)$.
Apply the quotient rule on the given function $y=\dfrac{x+1}{x-1}$while finding its derivative. So, since \[x+1\]is the first function and \[x-1\]is the second function then let\[f\left( x \right)=x+1\]and\[g\left( x \right)=x-1\]hence\[y=\dfrac{f\left( x \right)}{g\left( x \right)}\].
Differentiating equation (i) we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{g\left( x \right)\dfrac{df\left( x \right)}{dx}-f\left( x \right)\dfrac{dg\left( x \right)}{dx}}{{{\left( g\left( x \right) \right)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( x-1 \right)\dfrac{d\left( x+1 \right)}{dx}-\left( x+1 \right)\dfrac{d\left( x-1 \right)}{dx}}{{{\left( x-1 \right)}^{2}}}......\left( ii \right)\]
Differentiate the \[x+1\] and \[x-1\] in the equation (ii).
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( x-1 \right)\cdot 1-\left( x+1 \right)\cdot 1}{{{\left( x-1 \right)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{x-1-x-1}{{{\left( x-1 \right)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{2}{{{\left( x-1 \right)}^{2}}}......\left( iii \right)\]
Now we have got the equation (iii). In this equation we add \[-1\] and \[-1\] then subtract\[x\] and \[x\].
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{x-x-1-1}{{{\left( x-1 \right)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{2}{{{\left( x-1 \right)}^{2}}}\]
Thus, we have obtained the differential of the given function $y=\dfrac{x+1}{x-1}$ as \[\dfrac{dy}{dx}=-\dfrac{2}{{{\left( x-1 \right)}^{2}}}\].
Hence, the derivative of the function $y=\dfrac{x+1}{x-1}$ is \[-\dfrac{2}{{{\left( x-1 \right)}^{2}}}\].

Note:
Do not differentiate the functions directly using the quotient rule for such types of problems. The solution will be marked wrong if quotient rule is not applied to such questions.While differentiating a function we should always keep in mind that we know the formula for differentiation i.e. \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]. If two functions are in division form then we should never forget to apply the quotient rule of differentiation. There is no alternate way to solve this derivative because there is a function in the denominator. Keep in mind that: do not try to reduce the function so that it would not be in fraction form because it would make the function more complex.