Question

Let f(x) = tan(x) and let y^(n) denote the n^th derivative of f(x) then the value of y^(9998879879789776) is

(a) 908090988

(b) 0

(c) 989

(d) 1729

I got this question at a job interview.

Enquiry is from Leibniz Rule in section Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (b) 0

The explanation: Assume y = f(x) and we also know that tan(x)=\(\frac{sin(x)}{cos(x)}\)

Rewriting the function as y(cos(x))=sin(x)

Now differentiating on both sides upto n^nt derivative we have

(y(cos(x))^(n)=\(c_0^ny^{(n)}cos(x)-c_1^ny^{(n-1)}sin(x)+….+(cos(x))^{(n)}y\)

Now observe that y(0)=tan(0)=0….(1)

Now consider the second derivative at x=0 on both sides

(y(cos(x))^(2)=\(c_0^2y^{(2)}cos(0)-c_1^2y^{(1)}sin(0)-c_2^2ycos(0)=(sin(x))^{(2)}=0\)

Using (1) and the above equation one can conclude that

y^(2) = 0

This gives the value of second derivative to be zero

Similarly for any even value of n all the odd derivatives of y in the expression would have sin(x) as their coefficients and as the values of y^(0) and y^(2) are zero. Every even derivative of the tan(x) function has to be zero.

Thus, we have

y^(9998879879789776) = 0.