- Difficulty
- Capstone
- Suggested time
- 9 min
- Marks
- 5
The chain rule and logarithms practice question 29
Let be a differentiable function with for every , and . Let be the function below. Find .
Both the base and the exponent change with . Which method handles a variable base raised to a variable power?
Take logarithms first: . Differentiate the right side with the product rule, and with the chain rule, which here is used twice.
. Evaluate everything at , then multiply by .
An exact expression, or a decimal.
Theory UBC Math 100 §4.4CLP-1 §2.10Key results
Worked solution
Strategy
A variable base to a variable power calls for logarithmic differentiation. The logarithm brings the exponent down as a factor; the product rule and the chain rule, applied twice to log f(x²), do the rest, and only then are the values at x = 1 put in.
Working
The logarithm is defined because f > 0, and it brings the exponent down.
The product rule on the right; log f(x²) needs the chain rule twice, for log of f and for f of x².
At x = 1, x² = 1, so f(x²) = e and f′(x²) = 3.
Multiply back by g(1).
Answer
The answer is e plus 6, about 8.718.
Check
With f(t) = e · exp(3(t − 1)/e), which is positive everywhere and has f(1) = e and f′(1) = 3, the difference quotient (g(1.001) − g(0.999))/0.002 is about 8.7183.
Where this goes wrong
Using the power rule as if the exponent were fixed, which gives 6, or the rule for aˣ as if the base were fixed, which gives e. Both the base and the exponent vary, so logarithmic differentiation is needed, and the chain rule on f(x²) brings in the factor 2x.
Leans on
Laws of logarithms, from Functions, graphs and inverses
Try a similar question
Let . Use logarithmic differentiation to find .Saved in this browser only.