Question

we know that if ( u = f(x) ), then ( du = f(x)dx ). therefore, if ( u = x^{3}+38 ), then ( du=) ( dx ).

Explanation:

Step1: Differentiate \(u = x^{3}+38\)

Differentiate \(u = x^{3}+38\) with respect to \(x\).
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\) and \(\frac{d}{dx}(c)=0\) (where \(c\) is a constant).
For \(y=x^{3}+38\), \(\frac{du}{dx}=\frac{d}{dx}(x^{3}+38)\)

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Since \(du=\frac{du}{dx}dx\), then \(du = 3x^{2}dx\)

Answer:

\(3x^{2}\)