Question

Let f(x) = \(\sqrt{sin(x)}\) and let y^n denote the n^th derivative of f(x) at x = 0 then the value of the expression 12y^(5) y^(1) + 30 y^(4) y^(2) + 20 (y^(3))^2 is given by

(a) 0

(b) 655

(c) 999

(d) 1729

I had been asked this question during an interview.

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

Answer 1

Right answer is (a) 0

For explanation I would say: Assume y = f(x)

Rewriting the function as y^2 = sin(x)

Now applying Leibniz rule up to the sixth derivative we have

(y^2)^(6) = c0^6 y^(6) y + c1^6 y^(5) y(1) + ………+ c6^6 y^(6) y

(y^2)^(6) = 2 y^(6) y + 12 y^(6) y^(1) + 30 y^(4) y^(2) + 20 (y^(3))^2

(sin(x))^(6) = -sin(x)

Now substituting x = 0 and observing that y(0) = 0 we have

sin(0) = 0 = 12 y^(6) y^(1) + 30 y^(4) y^(2) + 20 (y^(3))^2.