If fnx-efn-1x - n - and f0x-x- then ddxfnx is equal to

F_n(x)=e^f_n 1(x) f_0(x)=x nbsp; nbsp; (given ) f_1(x)=e^x f_2(x)=e^e^x f_3(x)=e^e^e^x · · · f_n(x)=e^e ⋯ n times x ddx{f_n(x)}=f_n(x)ddxf_n 1(x) =f_n(x) ...

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Byju's Answer

Standard XII

Mathematics

Expansions of the Form (1+x)^(-n) and (1-x)^(-n)

If fnx=efn-1x...

Question

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

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

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

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

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n f_{n} (x)

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Solution

The correct option is C $f_{n} (x) \cdot f_{n - 1} (x) \dots f_{2} (x) \cdot f_{1} (x)$
$f_{n} (x) = e^{f_{n - 1} (x)}$
$f_{0} (x) = x$ $($ given $)$
$f_{1} (x) = e^{x}$
$f_{2} (x) = e^{e^{x}}$
$f_{3} (x) = e^{e^{e^{x}}}$
$\cdot$
$\cdot$
$\cdot$
$f_{n} (x) = e^{e \dots n t i m e s x}$
$\frac{d}{d x} {f_{n} (x)} = f_{n} (x)$ $\frac{d}{d x} f_{n - 1} (x)$
$= f_{n} (x) \cdot f_{n - 1} (x) \frac{d}{d x} f_{n - 2} (x)$
$\cdot$
$\cdot$
$\cdot$
$= f_{n} (x) \cdot f_{n - 1} (x) \cdot f_{n - 2} (x) \dots f_{1} (x) \cdot 1$

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Similar questions

Q. If

f_{n} (x) = e^{f_{n - 1} (x)}

for all

n ϵ N

and

f_{0} (x) = x

then

\frac{d}{d x} {f_{n} (x)}

is equal to

f_{n} (x) = e^{f_{n - 1} (x)}

for all

n ϵ N

and

f_{0} (x) = x

, then

\frac{d}{d x} {f_{n} (x)}

Q. 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))

Q. If

f_{0} (x) = \frac{x}{(x + 1)}

and

f_{n + 1} = f_{0} \circ f_{n} (x)

for

n = 0, 1, 2, \dots

then

f_{n} (x)

Q. Define a sequence

⟨ f_{0} (x), f_{1} (x), f_{2} (x), \dots ⟩

of functions by

f_{0} (x) = 1, f_{1} (x) = x, {(\begin{matrix} f_{n} (x) \end{matrix})}^{2} - 1 = f_{n + 1} (x) f_{n - 1} (x)

, for

n \geq 1

Prove that each

f_{n} (x)

is a polynomial with integer coefficients.

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Expansions of the Form (1+x)^(-n) and (1-x)^(-n)

Standard XII Mathematics

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If fn(x)=efn−1(x) ∀n∈N and f0(x)=x, then ddx{fn(x)} is equal to

The correct option is C fn(x)⋅fn−1(x)⋯f2(x)⋅f1(x)fn(x)=efn−1(x) f0(x)=x (given ) f1(x)=ex f2(x)=eex f3(x)=eeex ⋅ ⋅ ⋅ fn(x)=ee⋯n times x ddx{fn(x)}=fn(x)ddxfn−1(x) =fn(x)⋅fn−1(x)ddxfn−2(x) ⋅ ⋅ ⋅ =fn(x)⋅fn−1(x)⋅fn−2(x)⋯f1(x)⋅1

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