How do you find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ given $x^{2} y + x y^{2} = 6$ $x^2y+xy^2=6$ ?

Question

How do you find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ given $x^{2} y + x y^{2} = 6$ $x^2y+xy^2=6$ ?

2 Answers

Impact of this question

5123 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-30T06:35:18

$\frac{d^{2} y}{{d x}^{2}} = \frac{4 y (2 x + y) (x + y)}{x^{2} {(x + 2 y)}^{2}} - \frac{2 x y^{2} {(2 x + y)}^{2}}{{(x (x + 2 y))}^{3}} - \frac{2 y}{x (x + 2 y)}$ $(d^2y)/(dx^2)=(4y(2x+y)(x+y))/(x^2(x+2y)^2)-(2xy^2(2x+y)^2)/(x(x+2y))^3-(2y)/(x(x+2y)$

Explanation:

Generally differentiation problems involve functions i.e. $y = f (x)$ $y=f(x)$ - written explicitly as functions of $x$ $x$ . However, some functions y are written implicitly as functions of $x$ $x$ .

So what we do is to treat $y$ $y$ as $y = y (x)$ $y=y(x)$ and use chain rule. This means differentiating $y$ $y$ w.r.t. $y$ $y$ , but as we have to derive w.r.t. $x$ $x$ , as per chain rule, we multiply it by $\frac{d y}{d x}$ $(dy)/(dx)$ .

Here we are given $x^{2} y + x y^{2} = 6$ $x^2y+xy^2=6$

hence differentiating it we get

$2 x \times y + x^{2} \times \frac{d y}{d x} + x \times 2 y \frac{d y}{d x} + y^{2} = 0$ $2x xxy+x^2xx(dy)/(dx)+x xx2y(dy)/(dx)+y^2=0$

or $2 x y + x^{2} \frac{d y}{d x} + 2 x y \frac{d y}{d x} + y^{2} = 0$ $2xy+x^2(dy)/(dx)+2xy(dy)/(dx)+y^2=0$ ................(A)

i.e. $\frac{d y}{d x} = - \frac{2 x y + y^{2}}{x^{2} + 2 x y} = - \frac{y (2 x + y)}{x (x + 2 y)}$ $(dy)/(dx)=-(2xy+y^2)/(x^2+2xy)=-(y(2x+y))/(x(x+2y))$

Differentiating (A) further, we have

$2 x \frac{d y}{d x} + 2 y + x^{2} \frac{d^{2} y}{{d x}^{2}} + 2 x \frac{d y}{d x} + 2 y \frac{d y}{d x} + 2 x {(\frac{d y}{d x})}^{2} + 2 x y \frac{d^{2} y}{{d x}^{2}} + 2 y \frac{d y}{d x} = 0$ $2x(dy)/(dx)+2y+x^2(d^2y)/(dx^2)+2x(dy)/(dx)+2y(dy)/(dx)+2x((dy)/(dx))^2+2xy(d^2y)/(dx^2)+2y(dy)/(dx)=0$

or $\frac{d^{2} y}{{d x}^{2}} (x^{2} + 2 x y) + \frac{d y}{d x} (4 x + 4 y) + 2 x {(\frac{d y}{d x})}^{2} + 2 y = 0$ $(d^2y)/(dx^2)(x^2+2xy)+(dy)/(dx)(4x+4y)+2x((dy)/(dx))^2+2y=0$

or $\frac{d^{2} y}{{d x}^{2}} (x (x + 2 y)) = - 4 \frac{d y}{d x} (x + y) - 2 x {(\frac{d y}{d x})}^{2} - 2 y$ $(d^2y)/(dx^2)(x(x+2y))=-4(dy)/(dx)(x+y)-2x((dy)/(dx))^2-2y$

= $4 \frac{y (2 x + y)}{x (x + 2 y)} (x + y) - 2 x {(\frac{y (2 x + y)}{x (x + 2 y)})}^{2} - 2 y$ $4(y(2x+y))/(x(x+2y))(x+y)-2x((y(2x+y))/(x(x+2y)))^2-2y$

and $\frac{d^{2} y}{{d x}^{2}} = \frac{1}{x (x + 2 y)} (4 \frac{y (2 x + y)}{x (x + 2 y)} (x + y) - 2 x {(\frac{y (2 x + y)}{x (x + 2 y)})}^{2} - 2 y)$ $(d^2y)/(dx^2)=1/(x(x+2y))(4(y(2x+y))/(x(x+2y))(x+y)-2x((y(2x+y))/(x(x+2y)))^2-2y)$

= $\frac{4 y (2 x + y) (x + y)}{x^{2} {(x + 2 y)}^{2}} - \frac{2 x y^{2} {(2 x + y)}^{2}}{{(x (x + 2 y))}^{3}} - \frac{2 y}{x (x + 2 y)}$ $(4y(2x+y)(x+y))/(x^2(x+2y)^2)-(2xy^2(2x+y)^2)/(x(x+2y))^3-(2y)/(x(x+2y)$

Answer 2 · 2017-03-31T00:17:04

$\frac{d^{2} y}{{d x}^{2}} = \frac{6 x y (x^{2} + x y + y^{2}) (x + y)}{{(x^{2} + 2 x y)}^{3}}$ $(d^2y)/(dx^2) = {6xy(x^2+xy+y^2)(x+y) }/ (x^2+2xy)^3$

Explanation:

There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ $y$ explicitly as a function of $x$ $x$ , only implicitly through the equation $F (x, y) = 0$ $F(x, y) = 0$ which defines $y$ $y$ as a function of $x, y = y (x)$ $x, y = y(x)$ . Therefore we can write $F (x, y) = 0$ $F(x, y) = 0$ as $F (x, y (x)) = 0$ $F(x, y(x)) = 0$ . Differentiating both sides of this, using the partial chain rule gives us

$\frac{\partial F}{\partial x} (1) + \frac{\partial F}{\partial y} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$ $(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))$

And for the second derivative we similarly have:

$\frac{d y}{d x} = G (x, y) \Rightarrow \frac{d^{2} y}{{d x}^{2}} = \frac{\partial G}{\partial x} + \frac{\partial F}{\partial y} G (x, y)$ $dy/dx = G(x,y) => (d^2y)/(dx^2) = (partial G)/(partial x) + (partial F)/(partial y)G(x,y)$

So Let $F (x, y) = x^{2} y + x y^{2} - 6$ $F(x,y) = x^2y + xy^2 - 6$ ; Then;

$(partial F)/(partial x) = 2xy+y^2 \ \ \$ and $\ \ \ (partial F)/(partial y) = x^2+2xy$

And so:

$dy/dx = -(2xy+y^2)/(x^2+2xy)$

So to get the Second derivative;
Let $G(x,y) =-(2xy+y^2)/(x^2+2xy)$ ; Then:

$(partial G)/(partial x) = -{(x^2+2xy)(2y) - (2xy+y^2)(2x+2y)} / (x^2+2xy)^2$
$" " = 2y * {2x^2+3xy+y^2 - x^2-2xy} / (x^2+2xy)^2$
$" " = 2y * {x^2+xy+y^2 } / (x^2+2xy)^2$

$(partial G)/(partial y) = -{(x^2+2xy)(2x+2y) - (2xy+y^2)(2x)} / (x^2+2xy)^2$
$" " = 2x*{2xy+y^2-x^2-3xy-2y^2 } / (x^2+2xy)^2$
$" " = -2x*{xy+x^2+y^2 } / (x^2+2xy)^2$

And so using the above formula for the second derivative:

$(d^2y)/(dx^2) = (partial G)/(partial x) + (partial F)/(partial y)G(x,y)$

$" " = {2y(x^2+xy+y^2) } / (x^2+2xy)^2 + { 2x(xy+x^2+y^2)(2xy+y^2) } / {(x^2+2xy)^2(x^2+2xy)}$

Which we can simplify to get:

$(d^2y)/(dx^2) = {2y(x^2+xy+y^2)(x^2+2xy) + 2x(xy+x^2+y^2)(2xy+y^2) } / (x^2+2xy)^3$

$" " = {2xy(x^2+xy+y^2){ (x+2y) + (2x+y) } }/ (x^2+2xy)^3$

$" " = {6xy(x^2+xy+y^2)(x+y) }/ (x^2+2xy)^3$

Science

Math

Humanities

... and beyond

How do you find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ given $x^{2} y + x y^{2} = 6$ $x^2y+xy^2=6$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find (d^2y)/(dx^2)d2ydx2(d^2y)/(dx^2) given x^2y+xy^2=6x2y+xy2=6x^2y+xy^2=6?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ given $x^{2} y + x y^{2} = 6$ $x^2y+xy^2=6$ ?