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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.5.90

90. Find f'(0) for 
f(x) = e^(-1/x²), x≠0
   = 0, x = 0.

Guida verificata passo dopo passo
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Recognize that the function is defined piecewise: \( f(x) = e^{-1/x^2} \) for \( x \neq 0 \) and \( f(0) = 0 \). We need to find the derivative at \( x = 0 \), i.e., \( f'(0) \).
Recall the definition of the derivative at a point: \( f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \). Substitute the given function values to get \( f'(0) = \lim_{h \to 0} \frac{e^{-1/h^2} - 0}{h} = \lim_{h \to 0} \frac{e^{-1/h^2}}{h} \).
Analyze the behavior of the numerator \( e^{-1/h^2} \) as \( h \to 0 \). Since \( 1/h^2 \to \infty \), \( e^{-1/h^2} \to 0 \) very rapidly (faster than any polynomial rate).
Use this rapid decay to evaluate the limit \( \lim_{h \to 0} \frac{e^{-1/h^2}}{h} \). Consider whether the numerator approaches zero faster than the denominator approaches zero, which will determine the limit.
Conclude the value of \( f'(0) \) based on the limit evaluation, confirming whether the derivative exists and what its value is.

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The derivative of a function at a point measures the instantaneous rate of change or the slope of the tangent line at that point. It is defined as the limit of the difference quotient as the increment approaches zero, i.e., f'(a) = lim(h→0) [f(a+h) - f(a)] / h.
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