Question

Let $f(x)=\dfrac{({{x}^{2}}-x)}{({{x}^{2}}+2x)}$then $\dfrac{d\left[ {{f}^{-1}}\left( x \right) \right]}{dx}$ is equals to-
A. \[-\dfrac{3}{^{\mathop{\left( 1-x \right)}^{2}}}\]
B. \[\dfrac{3}{^{\mathop{\left( 1-x \right)}^{2}}}\]
C. \[\dfrac{1}{^{\mathop{\left( 1-x \right)}^{2}}}\]
D. \[-\dfrac{1}{^{\mathop{\left( 1-x \right)}^{2}}}\]

Answer 1

Hint: Type of questions are based on the function and differentiation. Which asks for both. As in this question one function is given, and we had to find out the differentiation of the inverse of the given function. So for this first we will find out the inverse of a given function, which states to find the inverse of function, we will replace ‘x’ with ‘y’ and ‘y’ with ‘x’ in the given function, and then find the value of y. And then we will simply differentiate it.

Complete step by step answer:
So, the given function in the question i.e. $f(x)=\dfrac{({{x}^{2}}-x)}{({{x}^{2}}+2x)}$, to find the inversed of given, we will replace ‘x’ with ;y; and ‘y’ with ‘x’ in whole given function. So the function we will get will be;
\[\begin{align}
  & f(x)=\dfrac{({{x}^{2}}-x)}{({{x}^{2}}+2x)} \\
 & y=\dfrac{({{x}^{2}}-x)}{({{x}^{2}}+2x)} \\
 & x=\dfrac{({{y}^{2}}-y)}{({{y}^{2}}+2y)} \\
\end{align}\]
Now we had to find out the value of ‘y’ from the above equation, so on solving we will get;
\[\begin{align}
  & x=\dfrac{({{y}^{2}}-y)}{({{y}^{2}}+2y)} \\
 & x=\dfrac{y(y-1)}{y(y+2)} \\
 & x=\dfrac{(y-1)}{(y+2)} \\
 & \dfrac{y+2}{y-1}=\dfrac{1}{x} \\
\end{align}\]
Now subtracting 1 from L.H.S and R.H.S side, we will get;
\[\begin{align}
  & \dfrac{y+2}{y-1}=\dfrac{1}{x} \\
 & \dfrac{y+2}{y-1}-1=\dfrac{1}{x}-1 \\
 & \dfrac{y+2-(y-1)}{y-1}=\dfrac{1-x}{x} \\
 & \dfrac{3}{y-1}=\dfrac{1-x}{x} \\
 & \dfrac{3x}{1-x}=y-1 \\
 & \dfrac{3x}{1-x}+1=y \\
 & \dfrac{2x+1}{1-x}=y \\
\end{align}\]
So, ${{f}^{-1}}\left( x \right)=\dfrac{2x+1}{1-x}$
Now, we had to differentiate it. Since it is in the$\dfrac{\text{I}}{\text{II}}$form, which can be differentiated by the quotient rule, which says that $\dfrac{d\left[ \dfrac{\text{I}}{\text{II}} \right]}{dx}=\dfrac{\text{II}\dfrac{d\left( \text{I} \right)}{dx}-\text{I}\dfrac{d\left( \text{II} \right)}{dx}}{^{\mathop{\left( \text{II} \right)}^{2}}}$, comparing it with our ${{f}^{-1}}\left( x \right)=\dfrac{2x+1}{1-x}$, $\text{I=2x+1}$and $\text{II=1-x}$. So now differentiating it with according to the formula we will get;
\[\begin{align}
  & \dfrac{d\left[ \dfrac{\text{I}}{\text{II}} \right]}{dx}=\dfrac{\text{II}\dfrac{d\left( \text{I} \right)}{dx}-\text{I}\dfrac{d\left( \text{II} \right)}{dx}}{^{\mathop{\left( \text{II} \right)}^{2}}} \\
 & \dfrac{d\left[ \dfrac{2x+1}{1-x} \right]}{dx}=\dfrac{\left( 1-x \right)\dfrac{d\left( 2x+1 \right)}{dx}-\left( 2x+1 \right)\dfrac{d\left( 1-x \right)}{dx}}{^{\mathop{\left( 1-x \right)}^{2}}} \\
 & \dfrac{d\left[ \dfrac{2x+1}{1-x} \right]}{dx}=\dfrac{\left( 1-x \right)(2)-\left( 2x+1 \right)(-1)}{^{\mathop{\left( 1-x \right)}^{2}}} \\
 & \dfrac{d\left[ \dfrac{2x+1}{1-x} \right]}{dx}=\dfrac{\left( 2-2x \right)-\left( -2x-1 \right)}{^{\mathop{\left( 1-x \right)}^{2}}} \\
 & \dfrac{d\left[ \dfrac{2x+1}{1-x} \right]}{dx}=\dfrac{2-2x+2x+1}{^{\mathop{\left( 1-x \right)}^{2}}} \\
 & \dfrac{d\left[ \dfrac{2x+1}{1-x} \right]}{dx}=\dfrac{3}{^{\mathop{\left( 1-x \right)}^{2}}} \\
 & \dfrac{d\left[ {{f}^{-1}}\left( x \right) \right]}{dx}=\dfrac{3}{^{\mathop{\left( 1-x \right)}^{2}}} \\
\end{align}\]
So, \[\dfrac{d\left[ {{f}^{-1}}\left( x \right) \right]}{dx}=\dfrac{3}{^{\mathop{\left( 1-x \right)}^{2}}}\]

So, the correct answer is “Option B”.

Note: To solve such a question you should very well know how to differentiate the basic function. And be patient to apply the quotient rule formula while differentiating, that in the denominator we will have a square of $\text{II}$which is $\text{1-x}$ in our case.