 ## D13 Partial differentiation Partial derivatives reveal how a function with many variables changes when you adjust just one of the variables in the input.

Let us suppose that we have the equation for a paraboloid with an elliptical cross-section such as $$z=x^{2}+4y^{2}.$$ In this case we have a function of two independent variables, $$z=f\left(x,y\right),$$ and its graph is a 3-dimensional surface. We need to be able to differentiate $$z$$ with respect to either $$x$$ or $$y$$. If we treat one of the variables, say $$y$$, as a constant, then we can treat $$z$$ as a function of just one variable, $$x$$. We can then calculate the derivative of $$z$$ with respect to $$x$$. This derivative is called the partial derivative of $$z$$ with respect to $$x$$ and is denoted by $$\frac{\partial z }{\partial x}$$. If we treat $$x$$ as a constant then we can treat $$z$$ as a function of $$y$$ and we can then calculate $$\frac{\partial z }{\partial y}$$ the partial derivative of $$z$$ with respect to $$y$$.

### Examples

1. If $$z=x^{2}+4y^{2}$$ find $$\frac{\partial z }{\partial x}$$ and $$\frac{\partial z }{\partial y}$$

\begin{align*} \frac{\partial z }{\partial x} & =2x+0\quad[\textrm{since y is treated as a constant}]\\ & =2x\\ \frac{\partial z }{\partial y} & =0+8y\quad[\textrm{since x is treated as a constant}]\\ & =8y. \end{align*}

1. If $$f(x,y)=xy+2y^{2}$$ find $$\frac{\partial f }{\partial x}$$ and $$\frac{\partial f }{\partial y}$$ \begin{align*} \frac{\partial f }{\partial x} & =y+0\quad[\text{Just as the derivative of 2x\text{ is x, the derivative of xy\text{ is y\text{ } } } }\\ & \quad\quad\quad\quad\quad\text{when y\text{ is treated as a constant]} }\\ & =y\\ \frac{\partial f }{\partial y} & =x+4y \end{align*}

2. If $$f(x,y)=x^{2}y+y^{2}\sin x$$ find $$\frac{\partial f }{\partial x}$$ and $$\frac{\partial f }{\partial y}$$ \begin{align*} \frac{\partial f }{\partial x} & =2xy+y^{2}\cos x\\ \frac{\partial f }{\partial y} & =x^{2}+2y\sin x \end{align*}

3. If $$f(x,y)=xy^{2}\sin(xy)$$ find $$\frac{\partial f }{\partial x}$$ and $$\frac{\partial f }{\partial y}$$ \begin{align*} \frac{\partial f }{\partial x} & =y^{2}\sin(xy)+xy^{2}\times y\cos(xy)[\textrm{applying the product rule}]\\ & =y^{2}\sin(xy)+xy^{3}\cos(xy)\\ \frac{\partial f }{\partial y} & =2xy\sin(xy)+xy^{2}\times x\cos(xy)\\ & =2xy\sin(xy)+x^{2}y^{2}\cos(xy) \end{align*}

### Alternative Notation and Evaluation at a Point

An alternative notation for partial derivatives is $$f_{x}$$ for $$\frac{\partial f }{\partial x}$$and $$f_{y}$$ for $$\frac{\partial f }{\partial y}$$.

Partial derivatives may be evaluated at particular points: $$f_{x}(2,1)$$ refers to the value of the partial derivative of $$f$$ with respect to $$x$$ at the point where $$x=2$$ and $$y=1$$.

#### Example

If $$f(x,y)=2x^{3}y+3y^{2}$$ find $$f_{y}(1,3)$$

\begin{align*} f(x,y) & =2x^{3}y+3y^{2}\\ \Rightarrow f_{y} & =2x^{3}+6y\\ \Rightarrow f_{y}(1,3) & =2+18\\ & =20. \end{align*}

### Higher Order Partial Derivatives

As we have seen, a function $$z$$= $$f(x,y)$$ has two partial derivatives. They are $$\frac{\partial f }{\partial x}$$ and $$\frac{\partial f }{\partial y}.$$

A function such as this will have four second order partial derivatives:

1. It can be differentiated with respect to $$x$$ and then with respect to $$x$$ again $$\frac{\partial }{\partial x}\left(\frac{\partial f }{\partial x}\right)$$=$$\frac{\partial ^2 f }{\partial x^{2}}=$$ $$f_{xx}$$

2. It can be differentiated with respect to $$y$$ and then with respect to $$y$$ again$$\frac{\partial }{\partial y}\left(\frac{\partial f }{\partial y}\right)$$=$$\frac{\partial ^ {2} f }{\partial y^2}=$$ $$f_{yy}$$

3. It can be differentiated with respect to $$x$$ and then with respect to $$y$$ $$\frac{\partial }{\partial y }\left(\frac{\partial f }{\partial x}\right)$$=$$\frac{\partial ^2 f }{\partial x\partial y}=$$ $$f_{xy}$$

4. It can be differentiated with respect to $$y$$ and then with respect to $$x$$ $$\frac{\partial }{\partial x}\left(\frac{\partial f }{\partial y}\right)$$=$$\frac{\partial ^2 f }{\partial y\partial x}=$$ $$f_{yx}$$

#### Example

Find all second order partial derivatives of $$f(x,y)=x^{2}y^{3}+2x\cos y$$

\begin{align*} f_{x} & =2xy^{3}+2\cos y\\ \Rightarrow f_{xx}=2y^{3} & \textrm{and }f_{xy}=6xy^{2}-2\sin y\\ f_{y} & =3x^{2}y-2x\cos y\\ \Rightarrow f_{yy}=6x^{2}y-2x\sin y & \textrm{and f_{yx} }=6xy^{2}-2\sin y \end{align*}

(Note that $$f_{xy}=f_{yx}.$$ This will always be the case.)

#### Exercises

$$1.$$ Find the partial derivatives $$\frac{\partial z }{\partial x}$$ and $$\frac{\partial z }{\partial y}$$for each of the following

$$\quad\text{a) z=3x^{4}+2y^{3} }$$

$$\quad$$b) $$z=x^{2}y$$

$$\quad\text{c)}$$ $$z=3xe^{2y}$$

$$\quad\text{d)}$$ $$z$$ = $$\ln(x^{3}y^{5}-2)$$

$$\quad\text{a) \frac{\partial z }{\partial x}=12x^{3}, \frac{\partial z }{\partial y} =6y^{2} }$$

$$\quad$$b) $$\frac{\partial z }{\partial x}=2xy$$, $$\frac{\partial z }{\partial y}=x^{2}$$

$$\quad\text{c)}$$ $$\frac{\partial z }{\partial x}=3e^{2y},\frac{\partial z }{\partial y}=6xe^{2y}$$

$$\quad\text{d)}$$ $$\frac{\partial z }{\partial x}$$=$$\frac{3x^{2}y^{5}}{x^{3}y^{5}-2},$$ $$\frac{\partial z }{\partial y}=\frac{5x^{3}y^{4}}{x^{3}y^{5}-2}$$

$$2.$$ Find the value of the indicated partial derivative at the given point

$$\text{\quad a)}$$ $$f(x,y)=x^{4}-4y^{2},\text{ find }f_{x}(2,3)$$

$$\text{\quad b})$$ $$f(x,y)=\ln(x^{2}+y^{3})\text{, find }f_{y}(-1,1)$$

$$\text{\quad a)}$$ 32

$$\text{\quad b})$$ 1.5

$$3.$$ Find the first and second order partial derivatives of the following

$$\quad\text{a)}$$ $$f(x,y)=x\ln(y)$$

$$\quad\text{b)}$$ $$f(x,y)=x^{3}+x^{2}y-3xy^{2}+y^{3}$$

$$\quad\text{c)}$$ $$f(x,y)=\sin(xy)$$

$$\quad\text{d)}$$ $$f(x,y)=x\cos y+ye^{x}$$

$$\quad\text{a)}$$ $$f_{x}=\ln y,$$ $$f_{y}=\frac{x}{y}$$, $$f_{xx}=0,f_{xy}=-\frac{x}{y^{2}}$$,$$f_{xy}=f_{yx}=\frac{1}{y}$$

$$\quad\text{b)}$$ $$f_{x}=3x^{2}+2xy-3y^{2},$$ $$f_{y}=x^{2}-6xy+3y^{2},$$

$$\qquad\ f_{xx}=6x+6y,f_{yy}=-6x+6y,f_{xy}=f_{yx}=2x-6y$$

$$\quad\text{c)}$$ $$f_{x}=y\cos(xy),\,f_{y}=x\cos(xy),\,f_{xx}=-y^{2}\sin(xy),$$

$$\quad\!\quad$$ $$f_{yy}=-x^{2}\sin(xy),\,f_{xy}=f_{yx}=-xy\sin(xy)+\cos(xy)$$

$$\quad\text{d)}$$ $$f_{x}=\cos y+ye^{x},$$ $$f_{y}=x\sin y+e^{x},$$ $$f_{xx}=ye^{x},$$

$$\quad\quad\!\ f_{yy}=-x\cos y,f_{xy}=f_{yx}=e^{x}-\sin y$$

What's next... D14 Higher Order Derivatives

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