Question

Let f(x) = sin(x)/1+x^2. Let y^(n) denote the n^th derivative of f(x) at x = 0 then the value of y^(100) + 9900y^(98) is

(a) 0

(b) -1

(c) 100

(d) 1729

The question was posed to me in an international level competition.

This interesting question is from Leibniz Rule topic in division Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (a) 0

For explanation: The key here is a simple manipulation and application of the Leibniz rule.

Rewriting the given function as

y(1 + x^2) = sin(x)…….(1)

The Leibniz rule for two functions is given by

(uv)^(n)=\(c_{0}^{n}u(v)^{(n)}+c_{1}^{n}u^{(1)}(v)^{(n-1)}+….+c_{n}^{n}u^{(n)}v\)

Differentiating  expression  (1) in accordance to Leibniz rule (upto the hundredth derivative) we have

(y(1+x^2))^(100) = \(c_{0}^{100}y^{(100)}(1+x^2)+c_{1}^{100}y^(99)(2x)+c_{2}^{100}y^(98)(2)+0….+0\)

(y(1+x^2))=(sin(x))^(100)=sin(x)

Now substituting gives us

y^(100)+9900y^(98)=sin(0)=0

Hence, Option 0 is the required answer.