If fₙ(x)=e^{fₙ-1}^{(x)}}, for all n N and f₀(x)=x, then d/dx \{fₙ(x)\}, is equal to

Correct Answer is : fn(x).fn−1(x)...f2(x)f1(x)fₙ(x). fₙ-1}(x) ... f₂(x) f₁(x)fn(x).fn−1(x)...f2(x)f1(x)

fₙ(x) · fₙ-1}(x) f₂(x) f₁(x)

fₙ(x). fₙ-1}(x) ... f₂(x) f₁(x)

Correct Answer is : fn(x).fn−1(x)...f2(x)f1(x)fₙ(x). fₙ-1}(x) ... f₂(x) f₁(x)fn(x).fn−1(x)...f2(x)f1(x)

Test Index

Manipal Entrance Test 2019 Math Paper

Question : 38 of 70

Marks: +1, -0

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

n\in N

f_{0}(x)=x,

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

is equal to

Solution:

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

=e^{f_{n-1}(x)} \frac{d}{dx} \{f_{n-1}(x)\}

=f_{n}(x) \cdot \frac{d}{dx} \{e^{f_{n-1}(x)}\}

=f_{n}(x) \cdot e^{f_{n-2}(x)} \cdot \frac{d}{dx} \{f_{n-2}(x)\}

=f_{n}(x) \cdot f_{n-1}(x) \dots f_{2}(x) \frac{d}{dx} \{f_{1}(x)\}

=f_{n}(x) f_{n-1}(x) \dots f_{2}(x) \cdot f_{1}(x) \cdot \frac{d}{dx} e^{f_{0}(x)}

=f_{n}(x) \cdot f_{n-1}(x) \dots f_{2}(x) \cdot f_{1}(x) \cdot \frac{d}{dx}(f_{0}(x))

=f_{n}(x) \cdot f_{n-1}(x) \dots f_{2}(x) \cdot f_{1}(x) \cdot \frac{d}{dx}(x)

[\because f_{0}(x)=x]

given

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

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

Go to Question:

Prev Question

Next Question