If fn-1x-lnfnx - n- N and f0x-x-1- then ddxfnx is

F_n 1(x)=ln (f_n(x)) Diffrentiating f'_n 1(x)=1f_n(x)× f'_n(x) f'_n(x)=f_n(x).f'_n 1(x) Replace n with n 1 f'_n 1(x)=f_n 1(x).f'_n 2(x) f'_n(x)=f_n(x).f_n 1(x) ...

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Byju's Answer

Standard XI

Mathematics

Chain Rule of Differentiation

If fn-1x=lnfn...

Question

If $f_{n - 1} (x) = ln (f_{n} (x)) \forall n \in N$ and $f_{0} (x) = x - 1$ , then $\frac{d}{d x} (f_{n} (x))$ is

f_{n} (x) \cdot f_{n - 1} (x)

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f_{n} (x) \cdot f_{n - 1} (x) \cdot f_{n - 2} (x) \dots f_{1} (x)

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f_{n - 1} (x) \cdot \frac{d}{d x} (f_{n - 1} (x))

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f_{n} (x) \cdot \frac{d}{d x} (f_{n - 1} (x))

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Solution

The correct option is D $f_{n} (x) \cdot \frac{d}{d x} (f_{n - 1} (x))$
$f_{n - 1} (x) = ln (f_{n} (x))$
Diffrentiating
$f_{n - 1}^{'} (x) = \frac{1}{f_{n} (x)} \times f_{n}^{'} (x)$
$f_{n}^{'} (x) = f_{n} (x) . f_{n - 1}^{'} (x)$
Replace $n$ with $n - 1$

$f_{n - 1}^{'} (x) = f_{n - 1} (x) . f_{n - 2}^{'} (x)$
$f_{n}^{'} (x) = f_{n} (x) . f_{n - 1} (x) . f_{n - 2} (x) . . . . . . f_{1} (x) . f_{0}^{'} (x)$ ....(i)
$f_{0} (x) = x - 1 \Rightarrow f_{0}^{'} (x) = 1$
From (i)
$f_{n}^{'} (x) = f_{n} (x) . f_{n - 1} (x) . f_{n - 2} (x) . . . . . . . f_{1} (x)$

Suggest Corrections

Similar questions

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If fn−1(x)=ln(fn(x)) ∀ n∈N and f0(x)=x−1, then ddx(fn(x)) is

If $f_{n - 1} (x) = ln (f_{n} (x)) \forall n \in N$ and $f_{0} (x) = x - 1$ , then $\frac{d}{d x} (f_{n} (x))$ is