Method of DifferentiationhardFree

Derivative of the inverse of h = g o f | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
If f(x)=x+1f(x) = x+1 and g−1(x)=x3+x+1g^{-1}(x) = x^3+x+1, let g∘f(x)=h(x)g\circ f(x) = h(x), then
lim⁡x→1h−1(x)−h−1(1)x−1=\lim_{x\to1}\frac{h^{-1}(x)-h^{-1}(1)}{x-1} =
Solution
Answer: 4
Step 1: By the definition of the derivative, the limit is
(h−1)′(1)\left(h^{-1}\right)'(1)
 Step 2: (g−1)′(x)=3x2+1>0\left(g^{-1}\right)'(x) = 3x^2+1>0, so g−1g^{-1}, and with it gg, is one-one, and h=g∘fh = g\circ f is invertible with
h=g∘f⟹h−1=f−1∘g−1h = g\circ f \quad\Longrightarrow\quad h^{-1} = f^{-1}\circ g^{-1}
 (undo gg first, then ff). Also
y=x+1  ⟹  x=y−1  ⟹  f−1(y)=y−1y = x+1 \;\Longrightarrow\; x = y-1 \;\Longrightarrow\; f^{-1}(y) = y-1
 Step 3:
h−1(x)=f−1(g−1(x))=(x3+x+1)−1=x3+xh^{-1}(x) = f^{-1}\left(g^{-1}(x)\right) = \left(x^3+x+1\right)-1 = x^3+x
 so gg itself is never needed. Step 4:
(h−1)′(x)=3x2+1⟹(h−1)′(1)=3+1=4\left(h^{-1}\right)'(x) = 3x^2+1 \quad\Longrightarrow\quad \left(h^{-1}\right)'(1) = 3+1 = 4
 Answer: 44
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