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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

Let ϕ ( x ) be the inverse of the function f ( x ) and f^ ′ (x) = 1...

Let ϕ(x)ϕ ( x ) be the inverse of the function f(x)andf′(x)=11+x5,f ( x ) and f^{′} \left(x\right) = \dfrac{1}{1 + x^{5}},then ddx\dfrac{d}{dx}ϕ\left(x\right) is equal to

Options

  1. A11+[ϕ(x)]5\dfrac{1}{1 + {\left[ϕ\left(x\right)\right]}^{5}}
  2. B11+[f(x)]5\dfrac{1}{1 + {\left[f\left(x\right)\right]}^{5}}
  3. C1+[ϕ(x)]51 + {\left[ϕ\left(x\right)\right]}^{5}
  4. D1+f(x)1 + f ( x )

Correct answer

C. 1+[ϕ(x)]51 + {\left[ϕ\left(x\right)\right]}^{5}

Step-by-step solution

We have, ϕ(x)=f−1(x)ϕ\left(x\right) = f^{- 1} ( x )

⇒x=f[ϕ(x)]⇒ x = f [ ϕ\left(x\right) ]

On differentiating both sides w.r.t. xx, we get

1=f′[ϕ(x)].ϕ′(x)1 = f^{'}\left[ϕ\left(x\right)\right]. {ϕ}^{'} ( x )

⇒ϕ′(x)=1f′[ϕ(x)]⇒ {ϕ}^{'}\left(x\right) = \dfrac{1}{f^{'}\left[ϕ\left(x\right)\right]}...(i)

Since, f′(x)=11+x5f^{′}\left(x\right) = \dfrac{1}{1 + x^{5}} (given)

f′(ϕ(x))=11+[ϕ(x)]5f^{'}\left(ϕ\left(x\right)\right) = \dfrac{1}{1 + \left[ϕ\left(x\right)\right]5}

From Equation (i),

{ϕ}^{'}\left(x\right) = \dfrac{1}{f^{'} [ ϕ\left(x\right) ]} = 1 + \left[ϕ\left(x\right)\right]5
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