Question

Let f(x) = \(\sqrt{1-x^2}\) and let y^(n) denote the n^th derivative of f(x) at x = 0 then the value of 6y ^(1) y^(2) + 2y^(3) is

(a) -998

(b) 0

(c) 998

(d) -1

The question was asked by my college director while I was bunking the class.

I would like to ask this question from Leibniz Rule in division Differential Calculus of Engineering Mathematics

Answer 1

The correct answer is (b) 0

Easiest explanation: Assume f(x) = y

Rewriting the function as

y^2 = 1 – x^2

Differentiating both sides of the equation up to the third derivative using leibniz rule we have

(y^2)^(3)=\(c_0^3y^{(3)}y+c_1^3y^{(2)}y^{(1)}+c_2^3y^{(1)}y^{(2)}+c_3^3y^{(3)}y\)

(y^2)^(3)=\(2y^{(3)}y+6y^{(2)}y^{(1)}\)

(1-x^2)^(3)=0

Now substituting x = 0 in both the equations and equating them yields

2y ^(3) y + 6y ^(2) y^(1) = 0.