- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
Functions of two variables and partial derivatives practice question 13
Let . Find the mixed second partial derivative , the derivative in of .
means differentiate in first, then in . At each stage, which factors of the function contain the variable you are differentiating in?
Find first. Both and contain , so use the product rule, with .
Then differentiate in , holding fixed: the product rule again. As a check, , and must agree.
A formula in x and y. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §15.2Key results
Worked solution
Strategy
Differentiate in x first, then differentiate the result in y. Both steps need the product rule, because exy contains both variables. The other order is shorter and makes a good check.
Working
Product rule in x; the chain rule gives the factor y from the exponent.
Product rule in y with x fixed: exy has y-derivative x exy, and 2x + x²y has y-derivative x².
Collect the terms and take out the common factor x².
The other order gives the same expression, as Clairaut’s theorem promises.
Answer
The answer is f sub x y equals x squared e to the x y, times (3 plus x y).
Check
For small y, f ≈ x² + x³y, so f_xy at y = 0 is 3x², and the formula gives x²·e⁰·3 = 3x² there.
Where this goes wrong
Differentiating x²exy in x as 2x exy, as if exy were a constant. It contains x, so f_x needs the product rule, and a slip in f_x carries straight into f_xy.
Leans on
The product rule, from Differentiation rules
Try a similar question
Let . Find , the derivative in of , and check it against .Saved in this browser only.