The chain rule and logarithms practice question 27
Let for . Use logarithmic differentiation to find .
Taking of both sides is the method asked for. What does it do to the exponent ?
Both the base and the exponent vary, so take logs first ( is the natural logarithm): .
Differentiate both sides: the chain rule makes the left side , and the right side needs the product rule. Then multiply through by .
A formula in x. 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 §4.4CLP-1 §2.10Key results
Worked solution
Strategy
Neither the power rule nor the exponential rule applies when the base and the exponent both vary. Taking logs brings the exponent down, the product rule differentiates what is left, and multiplying by f undoes the logarithm.
Working
The log of a power brings the exponent down as a factor; x > 0 keeps log x defined.
Differentiate both sides: the chain rule on the left, the product rule on the right.
Multiply both sides by f(x) = xsin x.
Answer
The answer is x to the sine x, times the quantity cos x log x plus sin x over x.
Check
At π/2 the exponent sin x is flat at its peak 1, so f behaves like x there, and the formula gives f′(π/2) = 1.
Where this goes wrong
Stopping at cos x log x + sin x/x. That is f′/f, the derivative of log f; multiply by f(x) = xsin x to get f′(x).
The idea behind it
The chain rule and logarithms lesson, a figure to move, in a few steps
Leans on
The product rule, from Differentiation rules
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