## - Partial Derivatives

For partial derivatives we differentiate with respect to one variable and treat the other variable(s) as constants.

$$f(x, y)$$ means that the function f contains two variables, namely $$x$$ and $$y$$.
$$f_x$$ means the first partial derivative with respect to $$x$$.
$$∂$$ is the symbol for partial derivatives.

First Partial Derivative
LESSON 1
Given that $$\displaystyle f(x, y)=x^3+2xy^2+{2x\over 5y}$$, evaluate
(i) $$f_x$$
(ii) $$f_y$$
SOLUTION
(i) For $$f_x$$ we differentiate $$f(x, y)$$ with respect to $$x$$, treating $$y$$ as a constant.
$$\displaystyle f(x,y)=x^3+2xy^2+{2\over 5y}x$$
$$\displaystyle f_x=3x^2+2y^2+{2\over 5y}$$
(ii) $$\displaystyle f(x,y)=x^3+2xy^2+{2x\over 5} y^{-1}$$
• Differentiate with respect to $$y$$, treating $$x$$ as a constant.
$$\displaystyle f_y=4xy-{2x\over5} y^{-2}$$
$$\displaystyle f_y=4xy-{2x\over 5y^2}$$

LESSON 2
Given that $$z=(3x^2+2xy+y^2)^4$$, determine
(i) $$\displaystyle {∂z\over ∂x}$$
(ii) $$\displaystyle {∂z\over ∂y}$$
SOLUTION
(i) $$z=(3x^2+2xy+y^2 )^4$$
$$\displaystyle {∂z\over ∂x}=4(3x^2+2xy+y^2)^3(6x+2y)$$
$$=8(3x+y) (3x^2+2xy+y^2)^3$$
(ii) $$z=(3x^2+2xy+y^2 )^4$$
$$\displaystyle {∂z\over ∂y}=4(3x^2+2xy+y^2 )^3 (2x+2y)$$
$$=8(x+y)(3x^2+2xy+y^2 )$$

LESSON 3
Given that $$f(r, s)=4r \ln⁡(r^2+s^2)$$, determine
(i) $$f_r$$
(ii) $$f_s$$
SOLUTION
(i) $$f(r,s)=4r \ln⁡(r^2+s^2 )$$
$$\displaystyle f_r=4 \ln⁡(r^2+s^2 )+4r\left({2r\over r^2+s^2}\right )$$
$$\displaystyle f_r=4 \ln⁡(r^2+s^2)+{8r^2\over r^2+s^2}$$

(ii) $$f(r, s)=4r \ln⁡(r^2+s^2)$$
$$\displaystyle f_s=4r \left({2s\over r^2+s^2}\right)$$
$$\displaystyle f_s={8rs\over r^2+s^2}$$

LESSON 4
Given that $$w=xy^2 z^3-y^4+e^z$$, determine
(i) $$\displaystyle {∂w\over ∂x}$$
(ii) $$\displaystyle {∂w\over ∂y}$$
(iii) $$\displaystyle {∂w\over ∂z}$$
SOLUTION
(i) $$w=xy^2 z^3-y^4+e^z$$
$$\displaystyle {∂w\over ∂x}=y^2 z^3$$

(ii) $$w=xy^2 z^3-y^4+e^z$$
$$\displaystyle {∂w\over ∂y}=2xyz^3-4y^3$$

(iii) $$w=xy^2 z^3-y^4+e^z$$
$$\displaystyle {∂w\over ∂z}=3xy^2 z^2+e^z$$

LESSON 5
For $$f(x, y)=x^3 y+2y$$, determine $$f_x (2,1)$$.
SOLUTION
$$f(x,y)=x^3 y+2y$$
• Differentiate partially with respect to $$x$$.
$$f_x=3x^2 y$$
• Substitute $$x=2$$ and $$y=1$$ into $$f_x$$
$$f_x (2, 1)=3(2)^2 (1)=12$$

Second Partial Derivative
LESSON 1
Given that $$\displaystyle f(x, y)=x^3 y-4xy^2+{y\over x^2}$$, determine
(i) $$f_{xx}$$
(ii) $$f_{xy}$$
(iii) $$f_{yx}$$
SOLUTION
(i) $$f(x,y)=x^3 y-4xy^2+yx^{-2}$$
• Determine the first partial derivative with respect to $$x$$.
$$f_x=3x^2 y-4y^2-2yx^{-3}$$
• Differentiate $$f_x$$ partially with respect to $$x$$
$$f_{xx}=6xy+{6\over x^4}$$

(ii) $$\displaystyle f(x,y)=x^3 y-4xy^2+{y\over x^2}$$
• Determine the first partial derivative with respect to $$x$$.
$$\displaystyle f_x=3x^2 y-4y^2-{2\over x^3 y}$$
• Differentiate $$f_x$$ partially with respect to $$y$$.
$$\displaystyle f_{xy}=3x^2-8y-{2\over x^3}$$

(iii) $$\displaystyle f(x,y)=x^3 y-4xy^2+{1\over x^2} y$$
• Determine the first partial derivative with respect to $$x$$.
$$f_y=x^3-8xy+x^{-2}$$
• Differentiate $$f_y$$ partially with respect to $$x$$
$$\displaystyle f_{yx}=3x^2-8y-{2\over x^3}$$
NB: $$f_{xy}=f_{yx}$$

LESSON 2
Given that $$\displaystyle w=x^4 y^2 z^2-\ln⁡(xy)-\sin⁡\left({2y\over 3z}\right)$$, determine
(i) $$\displaystyle {∂w\over ∂x}$$
(ii) $$\displaystyle {∂^2w \over ∂x∂y}$$
(iii) $$\displaystyle {∂w\over∂y}$$
(iv) $$\displaystyle {∂^2 w\over ∂y∂z}$$
(v) $$\displaystyle {∂w\over ∂z}$$
(vi) $$\displaystyle {∂^2 w\over ∂z^2}$$
SOLUTION
(i) $$\displaystyle w=x^4 y^2 z^2-\ln⁡(xy)-\sin⁡\left({2y\over 3z}\right)$$
$$\displaystyle {∂w\over ∂x}=4x^3 y^2 z^2-{1\over x}$$

(ii) We now have to differentiate $$\displaystyle {∂w\over ∂x}$$ partially with respect to $$y$$
$$\displaystyle {∂^2 w\over ∂x∂y}=8x^3 yz^2$$

(iii) $$\displaystyle w=x^4 y^2 z^2-\ln⁡(xy)-\sin⁡\left({2y\over 3z}\right)$$
$$\displaystyle {∂w\over ∂y}=2x^4 yz^2-{1\over y}-{2\over3z}\cos\left({2y\over 3z}\right)$$

(iv) $$\displaystyle {∂w\over ∂y}=2x^4 yz^2-{1\over y}-{\over 3} z^{-1} \cos⁡\left({2y\over 3} z^{-1}\right)$$

We now have to differentiate $$\displaystyle {∂w\over ∂y}$$ partially with respect to $$z$$

$$\displaystyle {∂^2 w\over ∂y∂z}=4x^4 yz+{2\over 3z^2} \cos\left({2y\over 3z}\right)-{2\over3z}\left[-\left(-{2y\over3z^2}\right)\sin\left({2y\over3z}\right)\right]$$
$$\displaystyle =4x^4 yz+{2\over 3z^2}\cos\left({2y\over3z}\right)-{4y\over9z^3}\sin\left({2y\over3z}\right)$$

(v) $$\displaystyle w=x^4 y^2 z^2-\ln⁡(xy)+\sin\left({2y\over 3}z^{-1}\right)$$
$$\displaystyle {∂w\over ∂z}=2x^4 y^2 z+\left(-{2y\over3z^2}\right)\cos\left({2y\over3z}\right)$$

$$\displaystyle =2x^4y^2z-{2y\over 3z^2}\cos \left({2y\over3z}\right)$$

(vi) $$\displaystyle {∂w\over ∂z}=2x^4 y^2 z-{2y\over 3} z^{-2} \cos⁡\left({2y\over 3} z^{-1} \right)$$

We now have to differentiate $$\displaystyle {∂w\over ∂z}$$ partially with respect to $$z$$

$$\displaystyle {∂^2 w\over ∂z^2}=2x^4 y^2+{4y\over 3z^3} \cos\left({2y\over3z}\right)-{2y\over3z^2}\left(-\left(-{2y\over3z^2}\right)\sin\left({2y\over3z}\right)\right)$$
$$\displaystyle =2x^4 y^2+{4y\over 3z^3}\cos\left({2y\over3z}\right)-{4y^2\over9z^4}\sin\left({2y\over3z}\right)$$