Functions of two variables and partial derivatives, question 14: UBC MATH 100 practice

Question 14 of 18Challenging3 marks · 6 minTheory UBC Math 100 §15.2Key results

Let $f(x, y) = \log\!\left(x^{2} + xy + 1\right)$ . Find $f_{xy}(2, 1)$ , the derivative in $y$ of $f_x$ , and check it against $f_{yx}(2, 1)$ .

Your answer

A fraction, or a decimal to four places.

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Theory UBC Math 100 §15.2Key results

Worked solution

Strategy

Differentiate in x first, then differentiate the result in y. Clairaut’s theorem says the other order gives the same answer, which makes a good check.

Working

$f_x = \frac{2x + y}{x^{2} + xy + 1}$
Chain rule on log: the derivative of the inside over the inside, with y held fixed.
$f_{xy} = \frac{(1)\left(x^{2} + xy + 1\right) - (2x + y)(x)}{\left(x^{2} + xy + 1\right)^{2}} = \frac{1 - x^{2}}{\left(x^{2} + xy + 1\right)^{2}}$
Quotient rule in y with x fixed: the top has y-derivative 1 and the bottom has y-derivative x.
$f_{xy}(2, 1) = \frac{1 - 4}{(4 + 2 + 1)^{2}} = -\frac{3}{49}$
Substitute x = 2 and y = 1.
$f_y = \frac{x}{x^{2} + xy + 1}, \qquad f_{yx} = \frac{(1)\left(x^{2} + xy + 1\right) - x(2x + y)}{\left(x^{2} + xy + 1\right)^{2}} = \frac{1 - x^{2}}{\left(x^{2} + xy + 1\right)^{2}}$
The other order gives the same expression, as Clairaut’s theorem promises.

Answer

f_{xy}(2, 1) = f_{yx}(2, 1) = -\frac{3}{49} \approx -0.0612

Check

With x = 2, f_x = (4 + y)/(5 + 2y) slips from 5/7 ≈ 0.7143 to about 0.7137 as y goes from 1 to 1.01, a slope near −0.06.

Where this goes wrong

Differentiating f_x in x a second time, which gives f_xx(2, 1) = −11/49. The subscripts give the order: f_xy is f_x differentiated in y.

The idea behind it

The functions of two variables and partial derivatives lesson, a figure to move, in a few steps

Leans on

Differentiating a logarithm, from The chain rule and logarithms

Try a similar question

Let

f(x, y) = \dfrac{x - y}{x + 2y}

. Find

\dfrac{\partial f}{\partial x}

(2, 1)

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Functions of two variables and partial derivatives practice question 14