# How do you find d2y/dx2 by implicit differentiation where x^2y + xy^2 = 3x?

Jun 3, 2018

$y ' ' = \frac{2 \cdot {x}^{4} y + 3 {x}^{3} {y}^{2} + 3 {x}^{2} {y}^{3} + x {y}^{4} + {x}^{3} y + 3 {x}^{2} {y}^{2} + 3 x {y}^{4} + 2 {y}^{4} - 3 {x}^{3} + 3 {x}^{2} y - 3 {x}^{2} - 9 x y - 6 {y}^{2} - 9 x}{{x}^{3} \left(x + 2 y\right) {\left(x + 2 x y\right)}^{2}}$

#### Explanation:

Differentiating
${x}^{2} y + x {y}^{2} = 3 x$
with respect to $x$ we get

$2 x y + {x}^{2} y ' + {y}^{2} + 2 x y y ' = 3$
so we get

$y ' = \frac{3 - {y}^{2} - 2 x y}{{x}^{2} + 2 x y}$

for the second derivative we obtain

$y ' ' = \frac{\left(- 2 y y ' - 2 y - 2 x y '\right) \left(x + 2 x y\right) - \left(3 - {y}^{2} - 2 x y\right) \left(2 x + 2 y + 2 x y '\right)}{{x}^{2} + 2 x y} ^ 2$

Now plug the result for $y '$ in this equation!