Partial Deriviatives
If \(f\) is a function of two variables \(x\) and \(y\), then
Partial derivative of \(f\) w.r.t \(x\) at \((a,b)\) is
\[f*x(a,b) = \lim*{h \rightarrow 0} \frac{f(a+h, b) - f(a,b)}{h}\]Partial derivative of \(f\) w.r.t \(y\) at \((a, b)\) is
\[f*y(a,b) = \lim*{h \rightarrow 0} \frac{f(a,b+h)-f(a,b)}{h}\]Notations for Partial Derivatives
If \(z=f(x,y)\), we write
\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}f(x,y) = \frac{\partial z}{\partial x} = f_x = D_xf=D_1f=f_1=f_x(x,y)\]Same for y.
Rules for finding Partial Derivatives of \(z=f(x,y)\)
[1] To find \(\frac{\partial f}{\partial x}\), regard \(y\) as a constant and differentiate \(f(x,y)\) w.r.t \(x\).
[2] To find \(\frac{\partial f}{\partial y}\), regard \(x\) as a constant and differentiate \(f(x,y)\) w.r.t \(y\).
Functions of more than two variables
If \(u\) is a function of \(n\) variables, \(u=f(x_1,...,x_n)\), then its partial derivative w.r.t. \(x_i\) is
\[\frac{\partial u}{\partial x*i} = \lim*{h \rightarrow 0} \frac{f(x_1,x_2,...,x_i+h,...,x_n)-f(x_1,...,x_n)}{h}\]Higher Derivatives
If \(f\) is a function of two variables, then its partial derivatives \(f_x\) and \(f_y\) are also functions of two variables so we can consider their partial derivatives \((f_x)_x, (f_x)_y, (f_y)_x, (f_y)_y\) which are called the second partial derivatives of \(f\). If \(z=f(x,y)\), then the second partial derivatives are
\[(f_x)\_y = \frac{\partial}{\partial y}(\frac{\partial f}{\partial x}) = \frac{\partial^2 f}{\partial y \partial x}\] \[(f_y)\_x = \frac{\partial}{\partial x}(\frac{\partial f}{\partial y}) = \frac{\partial^2 f}{\partial x \partial y}\]Clairaut’s Theorem
Suppose \(f\) is defined on a Disk \(D\) that contains the point \((a,b)\). If the function \(f_{xy}\) and \(f_{yx}\) are both continuous on \(D\), then
\[f*{xy}(a,b) = f*{yx}(a,b)\]Exercises
Q35
Find the first partial derivatives of the function
\[p = \sqrt{t^4 + u^2 \cos{v}}\] \[p_t = \frac{2t^3}{\sqrt{t^4+u^2\cos{v}}}\] \[p_u = \frac{u \cos{v}}{\sqrt{t^4+u^2\cos{v}}}\] \[p_v = \frac{-u^2 \sin{v}}{2\sqrt{t^4+u^2\cos{v}}}\]Q49
Use implicit differentiation to find \(\frac{\partial z}{\partial x}\)
\[e^z \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x} xy + yz\] \[\therefore \frac{\partial z}{\partial x} = \frac{yz}{e^z - xy}\]Q52
Find \(\frac{\partial z}{\partial x}\).
\[(a)\ z = f(x)g(y)\] \[(b)\ z=f(xy)\] \[(c)\ z = f(x/y)\]For (a),
\[\frac{\partial z}{\partial x} = f'(x)g(y)\]For (b),
\[\frac{\partial z}{\partial x} = f'(xy)y\]For (c),
\[\frac{\partial z}{\partial x} = \frac{f'(x/y)}{y}\]Q62
Verify that the conclusion of Clairaut’s Theorem hold: \(u_{xy} = u_{yx}\).
\[u = \ln(x+2y)\] \[u\_{xy} = -2(x+2y)^{-2}\] \[u\_{yx} = -2(x+2y)^{-2}\]Q80
If \(u = e^{a_1x_1+a_2x_2+ \cdots + a_nx_n}\), where \(a_1^2+a_2^2+ \cdots + a_n^2 = 1\), show that
\[\frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} + \cdots + \frac{\partial^2 u}{\partial x_n^2} = u\]Let’s find the second derivative of \(u\) w.r.t \(x_i\).
Also, let \(a_1x_1+a_2x_2+ \cdots + a_nx_n = T\).
\[\frac{\partial u}{\partial x_i} = e^{T}a_i\] \[\frac{\partial^2 u}{\partial x_i^2} = e^{T}a_i^2 = ua_i^2\]Hence,
\[\frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} + \cdots + \frac{\partial^2 u}{\partial x_n^2} = u(a_1^2 + a_2^2 + \cdots + a_n^2) = u\]Q97
You’re told that there is a function \(f\) whose partial derivatives are \(f_x(x,y) = x + 4y\) and \(f_y(x,y) = 3x - y\). Should you believe it?
Let’s compute \(f_{xy}\) and \(f_{yx}\)
\[f\_{xy} = 4\] \[f\_{yx} = 3\]By Clairaut’s theorem, since they’re different, such a function does not exist.
References
[1] Stewart Calculus 8th Edition