Exercise · ZW-DERIVATIVES-OF-EXPONENTIALS-AND-LOGARITHMS-007 · ★★☆ Intermediate

Differentiate \(f(x)=e^{x}\ln x\) for \(x>0\).

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  • D. \(e^{x}\ln x+\frac{e^{x}}{x}\)

Explanation

By the product rule: \((e^{x})'\ln x+e^{x}(\ln x)'=e^{x}\ln x+\frac{e^{x}}{x}\).

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Review the concept: Derivatives of exponentials and logarithms

The function \(e^x\) is its own derivative. For base \(a>0\), \(a^x\) has derivative \(a^x\ln a\), and \(\ln x\) has derivative \(\frac{1}{x}\) for \(x>0\).

\[(e^{x})'=e^{x},\quad(a^{x})'=a^{x}\ln a,\quad(\ln x)'=\frac{1}{x}\]

More practice problems: derivatives of exponentials and logarithms

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