Let f 1 x - e x - f 2 x - e f 1 x - - f n -1 x - e f n x - n - 1- The for any fixed n - d - dx f n x isA- f n x B- f n x f n -1 x C- f n x f n -1 x - f 1 x D- f n x - - f 1 x e x

The correct option is C f_n(x) f_n 1(x)...f_1(x)ddx f_n(x) = d e^f_n(x)dx=f_n(x).ddx( f_n(x)) = f_n(x).ddx( e^f_n 1(x)) =f_n(x)· f_n 1(x)·ddx( f_n 1(x)) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

One - One function

Let f 1 x = e...

Question

Let $f_{1} (x) = e^{x}, f_{2} (x) = e^{f_{1} (x)}, . . ., f_{n + 1} (x) = e^{f_{n} (x)} \forall n \geq 1.$ The for any fixed $n, \frac{d}{d x} f_{n} (x)$ is

f_{n} (x)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f_{n} (x) f_{n - 1} (x) . . . f_{1} (x)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f_{n} (x) . . . . . f_{1} (x) e^{x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $f_{n} (x) f_{n - 1} (x) . . . f_{1} (x)$
$\frac{d}{d x} f_{n} (x) = \frac{d e^{f_{n} (x)}}{d x} = f_{n} (x) . \frac{d}{d x} (f_{n} (x)) = f_{n} (x) . \frac{d}{d x} (e^{f_{n - 1} (x)}) = f_{n} (x) \cdot f_{n - 1} (x) \cdot \frac{d}{d x} (f_{n - 1} (x)) ⋮ ⋮ = f_{n} (x) f_{n - 1} (x) . . . f_{1} (x)$

Suggest Corrections

Similar questions

Q. Let

f_{1} (x) = e^{x}, f_{2} (x) = e^{f_{1} (x)}, . . . . . f_{n + 1} (x) = e^{f_{n} (x)}

for all

n \geq 1

. The for any fixed

n, \frac{d}{d x} 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.

Q. Let

f (x) = \frac{3}{4} x + 1,

and

f^{n} (x)

be defined as

f^{2} (x) = f (f (x))

and for

n \geq 2, f^{n + 1} (x) = f (f^{n} (x)) .

λ = lim n \to \infty f^{n} (x),

then

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. Let

f (x) = \frac{x + 1}{x - 1}

for all

x \neq 1

Let

f^{1} (x) = f (x), f^{2} (x) = f (f (x))

and generally

f^{n} (x) = f (f^{n - 1} (x))

for

n > 1

Let

P = f^{1} (2) f^{2} (3) f^{3} (4) f^{4} (5)

Which of the following is a multiple of P ?

Join BYJU'S Learning Program

Let f1(x)=ex,f2(x)=ef1(x),...,fn+1(x)=efn(x) ∀n≥1. The for any fixed n,ddxfn(x) is

The correct option is C fn(x)fn−1(x)...f1(x)ddxfn(x)=d efn(x)dx=fn(x).ddx(fn(x))=fn(x).ddx(efn−1(x))=fn(x)⋅fn−1(x)⋅ddx(fn−1(x)) ⋮ ⋮=fn(x)fn−1(x)...f1(x)

Let $f_{1} (x) = e^{x}, f_{2} (x) = e^{f_{1} (x)}, . . ., f_{n + 1} (x) = e^{f_{n} (x)} \forall n \geq 1.$ The for any fixed $n, \frac{d}{d x} f_{n} (x)$ is