Certainly! To find the derivative of the function [tex]\( y = x^3 + x^x + 1 \)[/tex], we need to apply the rules of differentiation. Let's break it down step-by-step:
1. Differentiate each term individually:
- [tex]\( y = x^3 \)[/tex]: Use the power rule for differentiation [tex]\(\frac{d}{dx}(x^n) = nx^{n-1}\)[/tex].
- [tex]\( y = x^x \)[/tex]: This term is more complex and requires using both the product rule and the chain rule.
- [tex]\( y = 1 \)[/tex]: The derivative of a constant is zero.
2. Differentiate [tex]\( x^3 \)[/tex]:
[tex]\[ \frac{d}{dx}(x^3) = 3x^2. \][/tex]
3. Differentiate [tex]\( x^x \)[/tex]:
- Rewrite [tex]\( x^x \)[/tex] as [tex]\( e^{x \ln(x)} \)[/tex].
- Let [tex]\( u = x \ln(x) \)[/tex], then [tex]\( x^x = e^u \)[/tex].
- Differentiate [tex]\( e^u \)[/tex] with respect to [tex]\( x \)[/tex], knowing that [tex]\( \frac{d}{dx}(e^u) = e^u \frac{du}{dx} \)[/tex].
- Differentiate [tex]\( u = x \ln(x) \)[/tex] using the product rule: [tex]\( \frac{d}{dx}(x \ln(x)) = \frac{d}{dx}(x) \cdot \ln(x) + x \cdot \frac{d}{dx}(\ln(x)) \)[/tex].
- [tex]\( \frac{d}{dx}(x) = 1 \)[/tex] and [tex]\( \frac{d}{dx}(\ln(x)) = \frac{1}{x} \)[/tex], thus:
[tex]\[ \frac{d}{dx}(x \ln(x)) = \ln(x) + 1. \][/tex]
- Combine these results to get the derivative of [tex]\( x^x \)[/tex]:
[tex]\[ \frac{d}{dx}(x^x) = x^x \cdot (\ln(x) + 1). \][/tex]
4. Differentiate the constant term [tex]\( 1 \)[/tex]:
[tex]\[ \frac{d}{dx}(1) = 0. \][/tex]
5. Combine all derivatives:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(x^3 + x^x + 1) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^x) + \frac{d}{dx}(1). \][/tex]
This yields:
[tex]\[ \frac{dy}{dx} = 3x^2 + x^x (\ln(x) + 1). \][/tex]
Thus, the derivative of the function [tex]\( y = x^3 + x^x + 1 \)[/tex] is:
[tex]\[ \boxed{3x^2 + x^x (\ln(x) + 1)}. \][/tex]
