Question

differentiate the function.

$y = \ln(e^{x}+xe^{x})$
$y=$

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differentiate the function.

$y = \tan(\ln(ax + b))$
$y=$

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Explanation:

Step1: Simplify the inner - function

First, factor out $e^{x}$ from $e^{x}+xe^{x}$, so $y = \ln(e^{x}(1 + x))=\ln(e^{x})+\ln(1 + x)=x+\ln(1 + x)$.

Step2: Differentiate term - by - term

The derivative of $x$ with respect to $x$ is $1$, and the derivative of $\ln(1 + x)$ using the chain - rule $\frac{d}{dx}\ln(u)=\frac{u'}{u}$ (where $u = 1 + x$ and $u'=1$) is $\frac{1}{1 + x}$.
So, $y'=1+\frac{1}{1 + x}=\frac{1 + x+1}{1 + x}=\frac{x + 2}{x + 1}$.

For the second function $y=\tan(\ln(ax + b))$:

Step1: Apply the chain - rule

Let $u=\ln(ax + b)$, then $y = \tan(u)$. The derivative of $y$ with respect to $u$ is $\sec^{2}(u)$, and the derivative of $u=\ln(ax + b)$ with respect to $x$ using the chain - rule $\frac{d}{dx}\ln(u)=\frac{u'}{u}$ (where $u = ax + b$ and $u'=a$) is $\frac{a}{ax + b}$.

Step2: Calculate the derivative of the whole function

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting the values, we get $y'=\sec^{2}(\ln(ax + b))\cdot\frac{a}{ax + b}=\frac{a\sec^{2}(\ln(ax + b))}{ax + b}$.

Answer:

For $y = \ln(e^{x}+xe^{x})$, $y'=\frac{x + 2}{x + 1}$; for $y=\tan(\ln(ax + b))$, $y'=\frac{a\sec^{2}(\ln(ax + b))}{ax + b}$