A) frac{-1}{{{(1-x)}^{2}}}\]
B) \[\frac{3}{{{(1-x)}^{2}}}\]
C) \[\frac{1}{{{(1-x)}^{2}}}\]
D) \[\frac{-3}{{{(1-x)}^{2}}}\]
Correct Answer: B
Solution :
Let \[y=\frac{{{x}^{2}}-x}{{{x}^{2}}+2x}\] \[\Rightarrow \] \[x=\frac{2y+1}{-y+1};\,x\ne 0\] \[\Rightarrow \] \[{{f}^{-1}}(x)=\frac{2x+1}{-x+1}\] \[\therefore \] \[\frac{d}{dx}\{{{f}^{-1}}(x)=\frac{(-x+1)\cdot 2-(2x+1)(-1)}{{{(-x+1)}^{2}}}\] \[=\frac{3}{{{(-x+1)}^{2}}}\]
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