Let φ(x) be the inverse of the function f(x) and f prime x=(1/1+x5), then (d/d x) φ x is equal to

Question

Let φ(x) be the inverse of the function f(x) and f prime x=(1/1+x5), then (d/d x) φ x is equal to (A) (1/1+φ x5) (B) (1/1+f x5) (C) 1+φ x5 (D) 1+f(x)

Tardigrade Admin · Accepted Answer

We have, φ x=f-1(x) ⇒ x=f[φ x] On differentiating both sides w.r.t. x, we get 1=f prime φ x ⋅ φ prime(x) ⇒ φ prime x=(1/f prime φ x) ldots text (i) Since, f prime x=(1/1+x5) (given) f prime φ x=(1/1+φ x5) From Equation (i), φ prime x=(1/f prime[φ x])=1+φ x5

Q. Let $ϕ (x)$ be the inverse of the function $f (x)$ and $f^{'} x = \frac{1}{1 + x ^{5}}$ , then $\frac{d}{d x} ϕ x$ is equal to

$\frac{1}{1 + ϕ x ^{5}}$

$\frac{1}{1 + f x ^{5}}$

$1 + ϕ x^{5}$

$1 + f (x)$

Q. Let ϕ(x) be the inverse of the function f(x) and f′x=1+x51​, then dxd​ϕx is equal to

1+ϕx51​

1+fx51​

1+ϕx5

1+f(x)

We have, ϕx=f−1(x) ⇒x=f[ϕx] On differentiating both sides w.r.t. x, we get 1=f′ϕx⋅ϕ′(x) ⇒ϕ′x=f′ϕx1​… (i) Since, f′x=1+x51​ (given) f′ϕx=1+ϕx51​ From Equation (i), ϕ′x=f′[ϕx]1​=1+ϕx5

Q. Let $ϕ (x)$ be the inverse of the function $f (x)$ and $f^{'} x = \frac{1}{1 + x ^{5}}$ , then $\frac{d}{d x} ϕ x$ is equal to

$\frac{1}{1 + ϕ x ^{5}}$

$\frac{1}{1 + f x ^{5}}$

$1 + ϕ x^{5}$

$1 + f (x)$