14.1 | The Partial Derivative


The partial derivative of \(f(x,y)\) with respect to \(x\) is given by
$$
f_x (x,y) = \lim_{h \rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h}
$$
and the partial derivative of \(f(x,y)\) with respect to \(y\) is given by
$$
f_y(x,y) = \lim_{h \rightarrow 0} \frac{f(x,y+h)-f(x,y)}{h}.
$$
The partial derivative of a function \(f(x,y)\) with respect to \(x\) is variously denoted by
$$
f_x(x,y), f_x, \frac{\partial f}{\partial x}, \frac{\partial}{\partial x} f(x,y), D_1 f, \text{ or } D_x f
$$
The method to compute higher order derivatives and the corresponding notation is suggested by the following:
$$
f_{xyxy}
$$
represents the 4th order derivative of \(f\) with respect to \(x\) first, \(y\) second, \(x\) third, and \(y\) fourth. That is,
$$
f_{xyxy} = \left( \left( \left( f_x \right)_y \right)_x \right)_y
$$

$$\frac{\partial^5 f}{\partial x \partial y \partial x^3}$$
represents the 5th order partial derivative of \(f\) with respect to \(x\) first, \(x\) second, \(x\) third, \(y\) fourth, and \(x\) fifth. That is,
$$
\frac{\partial^5 f}{\partial x \partial y \partial x^3} = \frac{\partial }{\partial x} \left( \frac{\partial }{\partial y}\left( \frac{\partial }{\partial x} \left( \frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right) \right)\right) \right)
$$

Clairaut’s Theorem: Suppose \(f(x,y)\) 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
$$
f_{xy}(a,b) = f_{yx}(a,b)
$$
Find the first partial derivatives of
$$
f(x,y) = x^2y+e^{\sin(x)}
$$


The first partial derivative of \(f\) with respect to \(x\) is \(2xy + \cos x e^{\sin(x)}\) and is obtained as follows:
$$
\begin{aligned}
\frac{\partial f}{\partial x} & = \frac{\partial }{\partial x}\left[ x^2y + e^{\sin(x)} \right] \\
& = \frac{\partial }{\partial x} \left[x^2 y \right] + \frac{\partial }{\partial x}\left[ e^{\sin(x)} \right] \\
& = y \cdot \frac{\partial }{\partial x}\left[ x^2 \right]+ \frac{\partial }{\partial x}\left[ e^{\sin(x)} \right] \\
& = 2xy + \frac{\partial }{\partial x}\left[ \sin(x) \right] \cdot e^{\sin(x)} \\
& = 2xy + \cos(x) e^{\sin(x)}
\end{aligned}
$$
where

  • in the second equality, we apply the sum rule
  • in the third equality, we apply the constant multiplier rule (since we are differentiating with respect to \(x\), \(y\) is considered constant)
  • in the fourth equality, we apply the power rule to the first term and chain rule to the second term
  • in the fifth equality, we apply a basic rule of differentiation: \(\frac{d}{dx}[\sin x] = \cos x\)

The first partial derivative of \(f\) with respect to \(y\) is \(2x\) and is obtained as follows:
$$
\begin{aligned}
\frac{\partial f}{\partial y} & = \frac{\partial }{\partial y}\left[ x^2y + e^{\sin(x)} \right] \\
& = \frac{\partial }{\partial y} \left[x^2 y \right] + \frac{\partial }{\partial y}\left[ e^{\sin(x)} \right] \\
& = x^2 \cdot \frac{\partial }{\partial y}\left[ y \right]+ \frac{\partial }{\partial y}\left[ e^{\sin(x)} \right] \\
& = 2x
\end{aligned}
$$
where

  • in the second equality, we apply the sum rule
  • in the third equality, we apply the constant multiplier rule (since we are differentiating with respect to \(y\), \(x\) is considered constant)
  • in the fourth equality, we apply the power rule to the first term and constant rule

E 14.1 Exercises