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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.A.15

15. Find f'(2) if f(x) = e^(g(x)) and g(x) = ∫(from 2 to x) t/(1+t⁴)dt.

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Identify the given functions: \( f(x) = e^{g(x)} \) and \( g(x) = \int_{2}^{x} \frac{t}{1+t^{4}} \, dt \). We need to find \( f'(2) \).
Recall that to find \( f'(x) \), we use the chain rule: \( f'(x) = e^{g(x)} \cdot g'(x) \). So, the next step is to find \( g'(x) \).
By the Fundamental Theorem of Calculus, the derivative of \( g(x) = \int_{2}^{x} \frac{t}{1+t^{4}} \, dt \) with respect to \( x \) is the integrand evaluated at \( x \): \( g'(x) = \frac{x}{1+x^{4}} \).
Substitute \( g'(x) \) back into the expression for \( f'(x) \): \( f'(x) = e^{g(x)} \cdot \frac{x}{1+x^{4}} \).
Finally, to find \( f'(2) \), evaluate \( g(2) = \int_{2}^{2} \frac{t}{1+t^{4}} \, dt \) (which is zero because the limits are the same), then substitute \( x=2 \) into \( f'(x) \) to get \( f'(2) = e^{g(2)} \cdot \frac{2}{1+2^{4}} \).

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Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if g(x) is defined as an integral with a variable upper limit, then g'(x) equals the integrand evaluated at x. Here, g(x) = ∫₂ˣ t/(1+t⁴) dt implies g'(x) = x/(1+x⁴).
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The chain rule is used to differentiate composite functions. For f(x) = e^(g(x)), the derivative f'(x) is e^(g(x)) multiplied by g'(x). This rule allows us to handle the exponential function with an inner function g(x).
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Evaluating Derivatives at a Point

After finding the general derivative f'(x), evaluating it at a specific point (x=2) involves substituting x=2 into both f(x) and g'(x). This step provides the exact slope of the function at that point.
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