Sure! Let's differentiate the function [tex]\( y = \ln(e^x + x e^x) \)[/tex].
To find [tex]\( y' \)[/tex], we'll need to use the chain rule and the properties of logarithms and exponential functions.
1. Identify the inner function:
[tex]\[ u = e^x + x e^x \][/tex]
2. Differentiate the natural logarithm:
According to the chain rule:
[tex]\[ y = \ln(u) \implies \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \][/tex]
3. Differentiate the inner function [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
u = e^x + x e^x
\][/tex]
Using the product rule for [tex]\( x e^x \)[/tex]:
[tex]\[
\frac{du}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(x e^x)
\][/tex]
The derivative of [tex]\( e^x \)[/tex] is [tex]\( e^x \)[/tex]:
[tex]\[
\frac{d}{dx}(e^x) = e^x
\][/tex]
To differentiate [tex]\( x e^x \)[/tex], use the product rule:
[tex]\[
\frac{d}{dx}(x e^x) = x \frac{d}{dx}(e^x) + e^x \frac{d}{dx}(x) = x e^x + e^x
\][/tex]
Combining these results:
[tex]\[
\frac{du}{dx} = e^x + (x e^x + e^x) = e^x + x e^x + e^x = x e^x + 2e^x
\][/tex]
4. Substitute [tex]\( \frac{du}{dx} \)[/tex] and [tex]\( u \)[/tex] into the chain rule expression:
[tex]\[
\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}
\][/tex]
[tex]\[
\frac{dy}{dx} = \frac{1}{e^x + x e^x} \cdot (x e^x + 2 e^x)
\][/tex]
5. Simplify the expression:
[tex]\[
\frac{dy}{dx} = \frac{x e^x + 2 e^x}{e^x + x e^x}
\][/tex]
Orthogonalizing terms we get:
[tex]\[
\frac{dy}{dx} = \frac{(x + 2) e^x}{(x + 1) e^x}
\][/tex]
6. Cancel out the common [tex]\( e^x \)[/tex] factor in the numerator and the denominator:
[tex]\[
\frac{dy}{dx} = \frac{x + 2}{x + 1}
\][/tex]
Hence, the derivative of the function [tex]\( y = \ln(e^x + x e^x) \)[/tex] is:
[tex]\[
y' = \frac{x e^x + 2 e^x}{x e^x + e^x}
\][/tex]
