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\( h ^ { \prime } ( x ) = e ^ { x } + \frac { 1 } { x ^ { 2 } } \)

Ask by Brewer Valdez. in Greece
Mar 15,2025

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Η αρχική συνάρτηση είναι \( h(x) = e^x - \frac{1}{x} + C \), όπου \( C \) είναι η σταθερά ολοκλήρωσης.

Solución

Αρχικά έχουμε τη παράγωγο \[ h'(x) = e^x + \frac{1}{x^2}. \] Για να βρούμε την αρχική συνάρτηση \( h(x) \), ολοκληρώνουμε κάθε όρο ως προς \( x \): 1. \(\displaystyle \int e^x \, dx = e^x\). 2. \(\displaystyle \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -\frac{1}{x}\). Συνεπώς, η ολοκληρωμένη συνάρτηση είναι \[ h(x) = e^x - \frac{1}{x} + C, \] όπου \( C \) είναι η αυθαίρετη σταθερά ολοκλήρωσης.

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Mind Expander

Did you know that derivatives like the one you’ve encountered have real-world applications in various fields? Take physics, for instance; they help us understand how things change over time! The derivative \( h'(x) \) indicates how the function \( h(x) \) changes with respect to \( x \). It’s like tracking the speed of a car—where the \( e^x \) part can signify continuous growth, and \( \frac{1}{x^2} \) highlights a diminishing influence as \( x \) increases! If you’re diving into derivatives and working with \( h'(x) \), be careful not to mix up exponential and polynomial rules! A common mistake is forgetting that the derivative of \( e^x \) is itself, while \( \frac{1}{x^2} \) falls under the power rule. Make sure to differentiate each term separately and check your algebra afterward—precision is your best friend in calculus!

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