Question

Find  dy/dx  for the function x³ + 2x²y - 3xy² + 5y³ = 27
1. \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3y^2-4xy- 3x^2}{2x^2-6xy+15y^2}\)
2. \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3y^2-4x- 3x^2}{2x^2-6xy+3y^2}\)
3. \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3y^2-4xy- 3x^2}{2x^2-6x+3y^2}\)
4. \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3y^2-4xy- 3x^2}{2x^2-6xy+3y^2}\)

Answer 1

Correct Answer - Option 1 : \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3y^2-4xy- 3x^2}{2x^2-6xy+15y^2}\)

Concept:

Derivative of xⁿ with respect to x is n x^n-1

Derivative of yⁿ with respect to x is n y^n-1 dy/dx.

Product rule: (uv)' = uv' + vu'

Calculation:

Given function is x3 + 2x2y - 3xy2 + 5y3 = 27

We differentiate the given function with respect to x

⇒ \(\rm 3x^2+ \frac{\mathrm{d} }{\mathrm{d} x}( 2x^2y )-\frac{\mathrm{d} }{\mathrm{d} x}(3xy^2)+15y^2\frac{\mathrm{d} y}{\mathrm{d} x}=0\)

As we know that, (uv)' = uv' + vu'

⇒ \(\rm 3x^2+ 2x^2\frac{\mathrm{d} y}{\mathrm{d} x} +4xy -6xy\frac{\mathrm{d} y}{\mathrm{d} x}-3y^2+15y^2\frac{\mathrm{d} y}{\mathrm{d} x}=0\)

⇒ \(\rm 3x^2+4xy-3y^2+ 2x^2\frac{\mathrm{d} y}{\mathrm{d} x} -6xy\frac{\mathrm{d} y}{\mathrm{d} x}+15y^2\frac{\mathrm{d} y}{\mathrm{d} x}=0\)

⇒ \(\rm 3x^2+4xy-3y^2+ (2x^2-6xy+15y^2)\frac{\mathrm{d} y}{\mathrm{d} x}=0\)

⇒ \(\rm (2x^2-6xy+15y^2)\frac{\mathrm{d} y}{\mathrm{d} x} = 3y^2-4xy- 3x^2\)

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3y^2-4xy- 3x^2}{2x^2-6xy+15y^2}\)

Hence, option 1 is correct.