Question

f(x, y)=\(\frac{x^3+y^3}{x^{99}+y^{98}x+y^{99}}\) find the value of fy at (x,y) = (0,1).

(a) 101

(b) -96

(c) 210

(d) 0

This question was addressed to me in an interview for job.

Question is from Euler’s Theorem topic in division Partial Differentiation of Engineering Mathematics

Answer 1

The correct option is (b) -96

Easy explanation: Using Euler theorem

xfx + yfy = n f(x, y)

Substituting x = 0; n=-96 and y = 1 we have

fy = -96. f(0, 1) = -96.(1⁄1)

= – 96.