Question

x^3 Sin(y) + Cos(x) y^3 = 0, its differentiation is?

(a) \(-\frac{[x^3 Sin(y)-3y^2 Sin(x)]}{[x^2 Cos(y)+y^3 Cos(x)]}\)

(b) \(-\frac{[3x^2 Sin(y)-y^3 Sin(x)]}{[x^3 Cos(y)+3y^2 Cos(x)]}\)

(c) \(-\frac{[3x^3 Sin(y)-y^3 Sin(x)]}{[x^3 Cos(y)+3y^3 Cos(x)]}\)

(d) 0

This question was addressed to me in exam.

I need to ask this question from Implicit Differentiation in chapter Partial Differentiation of Engineering Mathematics

Answer 1

Right option is (b) \(-\frac{[3x^2 Sin(y)-y^3 Sin(x)]}{[x^3 Cos(y)+3y^2 Cos(x)]}\)

Best explanation: \(\frac{d}{dx}[x^3 Sin(y)+Cos(x) y^3]\)= 0

\(3x^2 Sin(y)+x^3 Cos(y) \frac{dy}{dx}-Sin(x) y^3+3y^2 \frac{dy}{dx} Cos(x)=0\)

\(\frac{dy}{dx}=-\frac{[3x^2 Sin(y)-y^3 Sin(x)]}{[x^3 Cos(y)+3y^2 Cos(x)]}\)