This question aims to find all the second-order partial derivatives of the given function.
The derivative of a function with more than one variable with respect to one of the variables present in the function while treating the other variables as constant is called a partial derivative of that function. In other words, when the function input is composed of several variables, we are interested in seeing how the function changes when we change only a single variable while keeping the others constant. These types of derivatives are most commonly utilized in differential geometry and vector calculus.
The number of variables in a function remains the same when we take the partial derivative. Moreover, the higher order derivatives can be obtained by taking the partial derivatives of the already obtained partial derivatives. Higher-order derivatives are useful for determining the concavity of a function, that is, the maximum or minimum of a function. Let $f(x,y)$ be a function that is continuous and differentiable on an open interval, then two types of partial derivatives can be obtained namely direct second-order partial derivatives and cross partial derivatives, also known as mixed partial derivatives.
Expert Answer
First, partially differentiate $v$ with respect to $x$ keeping $y$ constant using the quotient rule as:
$v_x=\dfrac{(x-y)(y)-xy(1)}{(x-y)^2}$
$v_x=\dfrac{xy-y^2-xy}{(x-y)^2}$
$v_x=\dfrac{-y^2}{(x-y)^2}$
Second, partially differentiate $v$ with respect to $y$ keeping $x$ constant using the quotient rule as:
$v_y=\dfrac{(x-y)(x)-xy(-1)}{(x-y)^2}$
$v_y=\dfrac{x^2-xy+xy}{(x-y)^2}$
$v_y=\dfrac{x^2}{(x-y)^2}$
Now find the second-order partial derivatives and use the quotient rule as:
$v_{xx}=\dfrac{(x-y)^2(0)-(-y^2)[2(x-y)(1)]}{(x-y)^4}$
$v_{xx}=\dfrac{2y^2(x-y)}{(x-y)^4}$
$v_{xx}=\dfrac{2y^2}{(x-y)^3}$
$v_{yy}=\dfrac{(x-y)^2(0)-(x^2)[2(x-y)(-1)]}{(x-y)^4}$
$v_{yy}=\dfrac{2x^2(x-y)}{(x-y)^4}$
$v_{yy}=\dfrac{2x^2}{(x-y)^3}$
Also, find the mixed second-order partial derivatives as:
$v_{xy}=\dfrac{(x-y)^2(-2y)-(-y^2)[2(x-y)(-1)]}{(x-y)^4}$
$v_{xy}=\dfrac{-2y(x-y)^2-2y^2(x-y)}{(x-y)^4}$
$v_{xy}=\dfrac{2(x-y)[-y(x-y)-y^2]}{(x-y)^4}$
$v_{xy}=\dfrac{2[-xy+y^2-y^2]}{(x-y)^3}$
$v_{xy}=\dfrac{-2xy}{(x-y)^3}$
And it is well known that $v_{xy}=v_{yx}$.
Example 1
Let $f(x,y)=\sin(3x)+y^2e^{2x}-2x^2$ be a two-variable function. Find all the second-order partial derivatives of this function.
Solution
First, find the derivatives with respect to $x$ and $y$ as:
$f_x(x,y)=\cos(3x)\cdot 3+y^2\cdot (2e^{2x})-4x$
$f_x(x,y)=3\cos(3x)+2y^2e^{2x}-4x$
$f_y(x,y)=0+e^{2x}\cdot (2y)-0$
$f_y(x,y)=2ye^{2x}$
Now find the second order direct and mixed partial derivatives as:
$f_{xx}(x,y)=-3\sin(3x)\cdot 3+2y^2(2e^{2x})-4$
$f_{xx}(x,y)=-9\sin(3x)+4y^2e^{2x}-4$
$f_{yy}(x,y)=2e^{2x}$
$f_{xy}(x,y)=0+2(2y)e^{2x}-0$
$f_{xy}(x,y)=4ye^{2x}=f_{yx}(x,y)$
Example 2
Let $f(x,y)=ye^{xy^2}$. Prove that $f_{xy}=f_{yx}$.
Solution
The first order derivatives can be obtained as:
$f_x(x,y)=y(e^{xy^2}\cdot y^2)$
$f_x(x,y)=y^3e^{xy^2}$
$f_y(x,y)=y(e^{xy^2}\cdot 2xy)+e^{xy^2}\cdot 1$
$f_y(x,y)=2xy^2e^{xy^2}+e^{xy^2}$
$f_y(x,y)=e^{xy^2}(2xy^2+1)$
Now,
$f_{xy}(x,y)=y^3(2xye^{xy^2})+3y^2e^{xy^2}$
$f_{xy}(x,y)=2xy^4e^{xy^2}+3y^2e^{xy^2}$
$f_{xy}(x,y)=y^2e^{xy^2}(2xy^2+3)$ (1)
And,
$f_{yx}(x,y)=2xy^2(y^2e^{xy^2})+e^{xy^2}(2y^2)+y^2e^{xy^2}$
$f_{yx}(x,y)=2xy^4e^{xy^2}+2y^2e^{xy^2}+y^2e^{xy^2}$
$f_{yx}(x,y)=2xy^4e^{xy^2}+3y^2e^{xy^2}$
$f_{yx}(x,y)=y^2e^{xy^2}(2xy^2+3)$ (2)
So from equation (1) and (2) it is proves that $f_{xy}=f_{yx}$.
Example 3
Find $f_{xx}(x,y),f_{yy}(x,y)$ and $f_{xy}(x,y),f_{yx}(x,y)$ of the function $f(x,y)=x^2+y^2$.
Solution
The first order derivatives are:
$f_x(x,y)=2x+0$
$f_x(x,y)=2x$
$f_y(x,y)=0+2y$
$f_y(x,y)=2y$
The second-order derivatives are:
$f_{xx}(x,y)=2(1)$
$f_{xx}(x,y)=2$
$f_{yy}(x,y)=2(1)$
$f_{yy}(x,y)=2$
$f_{xy}(x,y)=0$
$f_{yx}(x,y)=0$