- 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 \(z\)

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