Let g(x) be the inverse of an invertible function f(x) which is differentiable at x = c , then g'(f(c)) equals

Question

Let g(x) be the inverse of an invertible function f(x) which is differentiable at x = c , then g'(f(c)) equals (A)  f'(c)  (B)  (1/f'(c))  (C)  f(c)  (D)  (1/f(c))

Tardigrade Admin · Accepted Answer

Since, g(x) is the inverse of an invertible function f(x). ∴ g[f(x)]=x ...(i) On differentiating Eq. (i) both sides w.r.t. ' x ', we get g prime[f(x)] f prime(x)=1 ⇒ g prime[f(x)]=(1/f prime(x)) ∴ g prime[f(c)]=(1/f prime(c))

Q. Let $g (x)$ be the inverse of an invertible function $f (x)$ which is differentiable at $x = c$ , then $g^{'} (f (c))$ equals

$f^{'} (c)$

$\frac{1}{f ^{'} ( c )}$

$f (c)$

$\frac{1}{f ( c )}$

Since, $g (x)$ is the inverse of an invertible function $f (x)$ .
$∴ g [f (x)] = x ... (i)$
On differentiating Eq. (i) both sides w.r.t. ' $x$ ', we get
$g^{'} [f (x)] f^{'} (x) = 1$
$\Rightarrow g^{'} [f (x)] = \frac{1}{f ^{'} ( x )}$
$∴ g^{'} [f (c)] = \frac{1}{f ^{'} ( c )}$

Q. Let g(x) be the inverse of an invertible function f(x) which is differentiable at x=c , then g′(f(c)) equals

f′(c)

f′(c)1​

f(c)

f(c)1​