Verified Answer :

olemakpadu

dy/dx = x²y²

d²y/dx² = is taking the derivative of the dy/dx

Note that if y = xⁿ,  dy/dx = nxⁿ ⁻ ¹.

For x²y² we will apply product rule of differentiation and implicit differentiation.

Product rule for y = uv

dy/dx = v(du/dx) + u(dv/dx)

dy/dx = x²y²

d²y/dx² = 2xy² + x²2y(dy/dx)

d²y/dx² = 2xy² + 2x²y(dy/dx)

But recall dy/dx = x²y² 

d²y/dx² = 2xy² + 2x²y(dy/dx)

d²y/dx² = 2xy² + 2x²y*( x²y²)

d²y/dx² = 2xy² + 2x⁴y³ = 2xy²(1 + x³y)

I hope this helps.

\(\frac{d^2y}{dx^2} =2xy^2+2x^4y^3\) will be the answer.

  Given derivative in the question,

  \(\frac{dy}{dx}=x^{2}y^{2}\)

To find the second derivative, differentiate the expression again by using partial differentiation,

\(\frac{d^2y}{dx^2}=\frac{dy}{dx}(x^2y^2)\)

     \(=y^2\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(y^2)\)

     \(=y^2(2x)+x^2(2y.\frac{dy}{dx})\)

     \(=2xy^2+2yx^2(\frac{dy}{dx})\)

     \(=2xy^2+2yx^2(x^2y^2)\)  [Since, \(\frac{dy}{dx}=x^{2}y^{2}\)]

     \(=2xy^2+2x^4y^3\)

  Therefore, \(\frac{d^2y}{dx^2}=2xy^2+2x^4y^3\) will be the answer.

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