Question

Let f(x) = e^e^x assuming all the n^th derivatives at x =0 to be 1 the value of the (n + 1)^th derivative can be written as

(a) e – 1 + 2^n

(b) 0

(c) 1

(d) None

I have been asked this question in an interview for internship.

I need to ask this question from Leibniz Rule topic in section Differential Calculus of Engineering Mathematics

Answer 1

Correct option is (a) e – 1 + 2^n

The explanation is: Assume y = f(x)

Taking ln(x) on both sides The function has to be written in the form ln(y) = e^x

Now computing the first derivative yields

 y^(1) = y * e^x

Now applying the Leibniz rule up to n^th derivative we have

y^(n+1)=\(c_0^ne^xy+c_1^ne^xy^{(1)}+….+c_n^ne^xy^{(n)}\)

We know that in the problem it is assumed that [y^(1)=y^(2)=…=y^(n)=1]x=0

Now, substituting x=0 we get

y^(n+1)=\(c_0^ne+c_1^n+….+c_n^n\)

From combinatorial results we know that 2^n=\(c_0^ne+c_1^n+….+c_n^n\)

This gives us

y^(n+1)=\(e+(c_0^ne+c_1^n+….+c_n^n)-c_0^n\)

y^(n+1)=e-1+2^n