# 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