If the function f is differentiable at x = c and is one-one in some neighbourhood of c, g is inverse function of f, then g'{f(c)} is:

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  1. \(\rm \frac{1}{f'(c)}\), where f'(c) ≠ 0
  2. g(c) f'(c) + g'(c) f(c)
  3. f'(c)
  4. \(\rm \frac{f(c)}{f'(c)}\) where f'(c) ≠ 0

Answer (Detailed Solution Below)

Option 1 : \(\rm \frac{1}{f'(c)}\), where f'(c) ≠ 0
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Detailed Solution

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Concept:

If g is the inverse function of f then gof(x) = g(f(x)) = x

Chain rule of differentiation: [g(f(x))]' = g'(f(x)).f '(x)

Calculation:

Given, the function f is differentiable at x = c and is one-one in some neighborhood of c, g is the inverse function of f,

⇒ g(f(x)) = x

Differentiating both sides with respect to x,

⇒ g'(f(x)).f '(x) = 1

⇒ g'(f(x)) = \(\rm \frac{1}{f'(x)}\)

So, at x = c,

g'(f(c)) = \(\rm \frac{1}{f'(c)}\)

∴ The correct option is (1).

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