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

##### 2 Answers

#### Explanation:

Generally differentiation problems involve functions i.e.

So what we do is to treat

Here we are given

hence differentiating it we get

or **................(A)**

i.e.

Differentiating (A) further, we have

or

or

=

and

=

# (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

# (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:

# dy/dx = G(x,y) => (d^2y)/(dx^2) = (partial G)/(partial x) + (partial F)/(partial y)G(x,y) #

So Let

#(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

# (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 #