let f(x)=x2-xx2+2x,x≠0,-2. Thenddxf-1(x) (wherever it is defined) is equal to:

f(x)=x2-xx2+2x f(x)=x-1x+2 f(x)=y ⇒ x=f-1(y) x-1x+2=y ⇒ x-1=xy+2y x(1-y)=2y+1 x=2y+11-y f-1(x)=2x+1-x+1 ddx(f-1(x))=3(1-x)2

let f(x)=x2-xx2+2x,x≠0,-2. Thenddxf-1(x) (wherever it is defined) is equal to:

f(x)=x2-xx2+2x f(x)=x-1x+2 f(x)=y ⇒ x=f-1(y) x-1x+2=y ⇒ x-1=xy+2y x(1-y)=2y+1 x=2y+11-y f-1(x)=2x+1-x+1 ddx(f-1(x))=3(1-x)2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

let f(x)= $\frac{x^{2} - x}{x^{2} + 2 x}$ ,x $\neq 0$ , $- 2$ . Then $\frac{d}{d x} [f^{- 1} (x)]$ (wherever it is defined) is equal to:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{- 1}{{(1 - x)}^{2}}$

$\frac{3}{{(1 - x)}^{2}}$

$\frac{1}{{(1 - x)}^{2}}$

$\frac{- 3}{{(1 - x)}^{2}}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = \frac{x^{2} - x}{x^{2} + 2 x} f (x) = \frac{x - 1}{x + 2} f (x) = y \Rightarrow x = f^{- 1} (y) \frac{x - 1}{x + 2} = y \Rightarrow x - 1 = xy + 2 y x (1 - y) = 2 y + 1 x = \frac{2 y + 1}{1 - y} f^{- 1} (x) = \frac{2 x + 1}{- x + 1} \frac{d}{dx} (f^{- 1} (x)) = \frac{3}{{(1 - x)}^{2}}$