5.4 Partial Derivatives
If \((x_0,y_0)\) is a point in the domain of a function \(z = f(x,y)\) then the derivative of \(z = f(x,y)\) with respect to \(x\) at \(x = x_0\) is defined as \[\dfrac {df}{dx}(x_0,y_0) = \dfrac {d f(x_0,y_0)}{dx} = \lim \limits _{\Delta x \rightarrow 0} \frac {f(x_0 + \Delta x, y_0) - (f(x_0,y_0)}{\Delta x}\] Provided that the limit exists. The limit is called the partial derivative of \(f\) with respect to \(x\) at the point \((x_0,y_0)\).
The usual notations of partial derivatives of \(z = f(x,y)\) with respect to \(x\) are \(\dfrac {d f(x_0,y_0)}{dx}\) or \(f_x(x_0,y_0)\) or \(\dfrac {dz}{dx}\Bigg |_{(x_0,y_0)}\)
\(f_x,\hspace {0.2cm} \dfrac {d f}{dx}\) or \(\dfrac {d z}{dx}\) is convenient when you regard the partial derivative as a function in its own right.
Note. \(\dfrac {\partial f}{\partial x}\) is the is the ordinary derivative of \(f\) with respect to \(x\) with \(y\) held fixed.
- 1.
- Find the value of \(\dfrac {\partial f}{\partial x}\) and \(\dfrac {\partial f}{\partial y}\) at \((4,5)\) if \(f(x,y) = x^2 + 3xy + y - 1\).
- 2.
- Find \(\dfrac {\partial f}{\partial y}\) if \(f(x,y) = y \sin xy\). \[\frac {\partial f}{\partial y} = \sin xy + yx\cos xy\]
- 3.
- Find \(f_x\) and \(f_y\) if \(f(x,y) = \dfrac {2y}{y + \cos x}\)
- 4.
- Find \(\dfrac {\partial [x\sin (y + 3z)]}{\partial z}\) \[\implies \hspace {0.5cm} \dfrac {\partial [x\sin (y + 3z)]}{\partial z} = 3x \cos (y + 3z)\]
Solution.
Part 1
\(\dfrac {\partial (x^2 + 3xy + y - 1)}{\partial x} = 2x + 3y\)
\(\dfrac {\partial f(4,5)}{\partial x} = 8 + 15 = 23\)
\(\dfrac {\partial f}{\partial y}= 3x + 1 \hspace {0.4cm} , \hspace {0.4cm} \dfrac {\partial f(4,5)}{\partial y} = 13\)
Part 3
\(f(x,y) = 2y(y + \cos x)^{-1}\)
\[f_x = \frac {\partial f}{\partial x} = -2y(y + \cos x)^{-2} (-\sin x) = \dfrac {2y\sin x}{(y + \cos x)^2}\]
\[f_y = \frac {\partial f}{\partial y} = \dfrac {2(y + \cos x) - 2y(1)}{(y + \cos x)^2} = \frac {2\cos x}{(y + \cos x)^2}\]
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