- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
The chain rule and logarithms practice question 22
Let . Use logarithmic differentiation to find .
Both the base and the exponent of depend on . Does the power rule or the rule for still apply, and what does a logarithm do to an exponent?
Neither rule applies when both the base and the exponent vary. Take of both sides first: .
Differentiate implicitly: . Then multiply back by .
An exact expression, or a decimal.
Theory UBC Math 100 §4.4CLP-1 §2.10Key results
Worked solution
Strategy
Take logarithms to move the variable exponent down as a multiplier, differentiate with the product and chain rules, then multiply through by the original function.
Working
x² + 1 > 0, so f(x) > 0 and log f(x) is defined. The log of a power brings the exponent down.
Differentiate both sides; the left is (log f)′ by the chain rule, the right by the product rule.
Evaluate at x = 1.
Multiply by f(1).
Answer
The answer is 2 natural log 2 plus 2, about 3.386.
Check
Holding the exponent at 1 gives a slope of 2, holding the base at 2 gives 2 log 2 ≈ 1.386, and the two effects add to 3.386.
Where this goes wrong
Treating the exponent as constant and writing x(x² + 1)x−1·2x, or treating the base as constant. With both varying, only logarithmic differentiation (or rewriting as ex log(x²+1)) is valid.
Leans on
Laws of logarithms, from Functions, graphs and inverses
Try a similar question
Let . Find .Saved in this browser only.