Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.A.15

Chapter 7, Problem 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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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).

Chain Rule

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).

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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