15.2 Higher order derivatives

Try writing $g=f_x$, and then writing the partial derivative of $g$ with respect to $y$.

From the example that “$f_x$" is the partial derivative of $f$ with respect to $x$, we infer that the notation for “take the partial derivative with respect to (variable)" is “write (variable) on the bottom right." Continuing this practice, to take the partial derivative with respect to $y$ of $f_x$, we should write the $y$ on the bottom right – that is, to the right of the $x$:

Since $x$ is to the left of $y$, we write the above as $f_{xy}$, not $f_{yx}$.

From the example that “$f_x$" is the partial derivative of $f$ with respect to $x$, we infer that the notation for “take the partial derivative with respect to (variable)" is “write (variable) on the bottom right." Continuing this practice, to take the partial derivative with respect to $y$ of $f_x$, we should write the $y$ on the bottom right – that is, to the right of the $x$:

Since $x$ is to the left of $y$, we write the above as $f_{xy}$, not $f_{yx}$.

Try writing $g=\pdiff{f}{x}$, and then writing the partial derivative of $g$ with respect to $y$.
You aren't asked about the power of the $f$; only the order of $x$ and $y$.

From the example that “$\pdiff{}{x}f$" is the partial derivative of $f$ with respect to $x$, we infer that the notation for “take the partial derivative of a function with respect to (variable)" is “put the partial derivative operator $\pdiff{}{(\text{variable})}$ to the left of the function." Continuing this practice, to take the partial derivative with respect to $y$ of $\pdiff{f}{x}$, we should write the operator $\pdiff{}{y}$ on the left.

In the above expression, $\partial y$ is to the left of the $\partial x$. So we write $\frac{\partial^2 f}{\partial y \partial x}$ rather than $\frac{\partial^2 f}{\partial x \partial y}$.

From the example that “$\pdiff{}{x}f$" is the partial derivative of $f$ with respect to $x$, we infer that the notation for “take the partial derivative of a function with respect to (variable)" is “put the partial derivative operator $\pdiff{}{(\text{variable})}$ to the left of the function." Continuing this practice, to take the partial derivative with respect to $y$ of $\pdiff{f}{x}$, we should write the operator $\pdiff{}{y}$ on the left.

In the above expression, $\partial y$ is to the left of the $\partial x$. So we write $\frac{\partial^2 f}{\partial y \partial x}$ rather than $\frac{\partial^2 f}{\partial x \partial y}$.

Look at the displayed equation in the answer to Question 2.

As in Question 2, if we want to differentiate $\pdiff{f}{x}$ with respect to $x$, we write:

In both cases:

As in Question 2, if we want to differentiate $\pdiff{f}{x}$ with respect to $x$, we write:

In both cases:

$\diff{}{x}[\tan x] = \sec^2 x$, $\diff{}{x}[\sec x] = \sec x \tan x$

see solution

We've separated out factors only depending on $x$, since these will act as constants when we differentiate with respect to $y$. Differentiating $\sec^2(xy)$ involves two layers of chain rule, so we'll figure that out on its own before we find $f_{xy}$.

Now we differentiate $f_x$ with respect to $y$.

To find $f_{yx}$, we first differentiate $f$ with respect to $y$.

We can differentiate this using the quotient rule. When we do, we'll need to find the derivative of the numerator. Since that takes several steps, we do it first.

We already found $\pdiff{}{x}\left[\sec^2(xy)\right]=2x\sec^2(xy)\cdot \tan(xy)$. By an equivalent calculation, $\pdiff{}{y}\left[\sec^2(xy)\right]=2y\sec^2(xy)\cdot \tan(xy)$

Now, let's differentiate $f_y$ with respect to $x$.

To show that this is equal to $f_{xy}$, we rearrange.

Save yourself time by using Theorem 15.2.5.

(a) $f_{xx}(x,y) = 2y^3$ $f_{yxy}(x,y) = f_{xyy}(x,y) = 12xy$

(b) $f_{xx}(x,y)= y^4e^{xy^2}$ $f_{xy}(x,y)= \big(2y+2xy^3\big)e^{xy^2}$ $f_{xxy}(x,y)= \big(4y^3 + 2xy^5\big)e^{xy^2}$ $f_{xyy}(x,y) = \big(2+10xy^2+4x^2y^4\big)e^{xy^2}$

(c) $\displaystyle\frac{\partial^3 f}{\partial w\,\partial v\,\partial u}(u,v,w) = -\frac{36}{(u+2v+3w)^4}$ $\displaystyle\frac{\partial^3 f}{\partial w\,\partial v\,\partial u}(3,2,1) = -0.0036 = -\frac{9}{2500}$

(a) We have

(b) We have

(c) We have

In particular,

Remember there are four second partial derivatives: $f_{xx}$, $f_{xy}$, $f_{yx}$, and $f_{yy}$.

$f_{xx}=\frac{5y^2}{(x^2+5y^2)^{3/2}}$ $f_{xy}=f_{yx}=-\frac{5xy}{(x^2+5y^2)^{3/2}}$ $f_{yy}=\frac{5x^2}{(x^2+5y^2)^{3/2}}$

Let $f(x,y)=\sqrt{x^2+5y^2}$. Then

Simplifying, and in particular using that $\frac{1}{\sqrt{x^2+5y^2}} =\frac{x^2+5y^2}{(x^2+5y^2)^{3/2}}$,

(a) This higher order partial derivative can be evaluated extremely efficiently by carefully choosing the order of evaluation of the derivatives.

(b) This higher order partial derivative can be evaluated extremely efficiently by carefully choosing a different order of evaluation of the derivatives for each of the three terms.

(c) Set $g(x) = f(x,0,0)$. Then $f_{xx}(1,0,0)=g''(1)$.

(a) $f_{xyz}(x,y,z)=0$ (b) $f_{xyz}(x,y,z)=0$ (c) $f_{xx}(1,0,0)=0$

(a) As $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z$, we have $f_z(x,y,z) = 0$ and hence

(b) Write $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$, $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ and $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$. Then

As $f(x,y,z)=u(x,y,z)+v(x,y,z)+w(x,y,z)$, we have

(c) In the course of evaluating $f_{xx}(x,0,0)$, both $y$ and $z$ are held fixed at $0$. Thus, if we set $g(x) = f(x,0,0)$, then $f_{xx}(x,0,0)=g''(x)$. Now

for all $x$. So $g'(x)=0$ and $g''(x)=0$ for all $x$. In particular,

See the solution.

As

we have

A similar method as Question 3 in Section 15.1, but iterated.

$f_{xy}(1.8,2.0)\approx0$

The definition of the derivative involves a limit at $h$ goes to 0; we can approximate that limit by choosing a value of $h$ that's close to 0; in our case, $0.1$ or $-0.1$ are the best we can do, using the information on the table.

The same holds for the second derivative:

These are the ideas we'll use in the approximations below.

The second partial derivative $f_{xy}(x,y)$ of $f$ is the partial derivative of $f_x(x,y)$ with respect to $y$. That is:

For our approximation, we can choose $h=0.1$ or $h=-0.1$. There's no compelling reason to choose one over the other. Let's use $h=0.1$.

Once again, there's no compelling reason to choose $h=0.1$ over $h=-0.1$. We could even choose different signs for the two limits. We'll just choose $h=0.1$ again, because after all, we do have to choose something.

Remark: different choices of $h$ all end up with the same approximation.