Question

How do you find the derivative of \[y=\ln {{\left( \dfrac{x-1}{x+1} \right)}^{\dfrac{1}{3}}}\]?

Answer 1

Hint: We will solve this equation using the concept of Quotient Rule. First we will rewrite the given equation by changing ln to base e. Then we will apply the quotient rule to the obtained equation. After applying quotient rules we have to simplify the obtained equation to obtain the result.

Complete step by step answer:
We have know about Quotient rule before solving the problem
Quotient rule:
In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Let \[f\left( x \right)=\dfrac{g\left( x \right)}{h\left( x \right)}\] where both \[g\] and \[h\] are differentiable and \[h\left( x \right)\ne 0\] The quotient rule states that the derivative of \[f\left( x \right)\] is
\[f'\left( x \right)=\dfrac{g'\left( x \right)h\left( x \right)-g\left( x \right)h'\left( x \right)}{{{\left[ h\left( x \right) \right]}^{2}}}\]
Now we will start solving our equation.
Given equation is
\[y=\ln {{\left( \dfrac{x-1}{x+1} \right)}^{\dfrac{1}{3}}}\]
Now we can rewrite the equation as follows as it is a ln function
\[\Rightarrow y=\dfrac{1}{3}\ln \left( \dfrac{x-1}{x+1} \right)\]
\[\Rightarrow 3y=\ln \left( \dfrac{x-1}{x+1} \right)\]
We have rule \[{{e}^{\ln \left( x \right)}}=x\] from that we can write \[{{e}^{x}}=\ln \left( x \right)\]
Using \[{{e}^{x}}=\ln \left( x \right)\] we can rewrite our equation as
\[\Rightarrow {{e}^{3y}}=\left( \dfrac{x-1}{x+1} \right)......\left( 1 \right)\]
Now we have to apply quotient rule to the equation
 \[f'\left( x \right)=\dfrac{g'\left( x \right)h\left( x \right)-g\left( x \right)h'\left( x \right)}{{{\left[ h\left( x \right) \right]}^{2}}}\]
From our equation
\[f'\left( x \right)=3{{e}^{3y}}\dfrac{dy}{dx}\]
\[g'\left( x \right)=1\]
\[h'\left( x \right)=1\]
So quotient rule will give the equation like
\[\Rightarrow 3{{e}^{3y}}\dfrac{dy}{dx}=\dfrac{\left( x+1 \right)\left( 1 \right)-\left( x-1 \right)\left( 1 \right)}{{{\left( x+1 \right)}^{2}}}\]
Now solving the equation
\[\Rightarrow 3{{e}^{3y}}\dfrac{dy}{dx}=\dfrac{x+1-x+1}{{{\left( x+1 \right)}^{2}}}\]
Now adding the similar terms we get
\[\Rightarrow 3{{e}^{3y}}\dfrac{dy}{dx}=\dfrac{2}{{{\left( x+1 \right)}^{2}}}\]
Now from equation 1 we can substitute \[\dfrac{x-1}{x+1}\] in place of \[{{e}^{3y}}\] so we get
\[\Rightarrow 3\left( \dfrac{x-1}{x+1} \right)\dfrac{dy}{dx}=\dfrac{2}{{{\left( x+1 \right)}^{2}}}\]
Now we have to rearrange the like terms we will get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{3}\times \dfrac{1}{{{\left( x+1 \right)}^{2}}}\times \dfrac{x+1}{x-1}\]
Simplifying the above equation we can cancel \[x+1\] present in both numerator and denominator
We will get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{3}\times \dfrac{1}{x+1}\times \dfrac{1}{x-1}\]
Now we will multiply the terms we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{3}\times \dfrac{1}{{{x}^{2}}-1}\]
So the derivative we will obtained is
\[\dfrac{dy}{dx}=\dfrac{2}{3{{x}^{2}}-3}\]

Hence the derivative of the given equation is
 \[y=\ln {{\left( \dfrac{x-1}{x+1} \right)}^{\dfrac{1}{3}}}=\dfrac{dy}{dx}=\dfrac{2}{3{{x}^{2}}-3}\]

Note:
We can also solve this problem without applying the quotient rule by simple derivatives. In this method we have to apply derivatives to ln function directly without converting to base e. For this we have to be aware of the formulas of derivatives of all the terms given in the equation.