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Manipal MET2019MathematicsDifferentiation

If f_n(x)=e^ f_ (n-1) (x) , for all n N and f₀(x)=x , then d d x f_n(x) is equal to

Options

  1. Af_n(x) f_ n-1 (x)
  2. Bf_n(x) d d x f_ n+1 (x)
  3. Cf_n(x) f_ n-1 (x) f₂(x) f₁(x)
  4. DNone of the above

Correct answer

C. f_n(x) f_ n-1 (x) f₂(x) f₁(x)

Step-by-step solution

d d x f_n(x) = d d x e^ f_ n-1 (x) =e^ f_ n-1 (x) d d x f_ n-1 (x) =f_n(x) d d x e^ f_ n-1 (x) =f_n(x) e^ f_ n-2 (x) d d x f_ n-2 (x) =f_n(x) f_ n-1 (x) f₂(x) d d x f₁(x) =f_n(x) f_ n-1 (x) f₂(x) f₁(x) d d x e^ f₀(x) =f_n(x) f_ n-1 (x) f₂(x) f₁(x) d d x (f₀(x) )=f_n(x) f_ n-1 (x) f₂(x) f₁(x) d d x (x) [ f₀(x)=x (given) ]=f_n(x) f_ n-1 (x) f₂(x) f₁(x) .

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