How do you find the derivative of $e^{1 + ln x}$ $e^(1+lnx)$ ?

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Answer 1 · 2017-06-04T00:33:40

$\frac{d}{d x} (e^{1 + ln x}) = e$ $d/dx(e^(1+lnx))=e$

We need $\frac{d}{d x} (e^{1 + ln x})$ $d/dx(e^(1+lnx))$

First, rewrite the expression using $a^{b + c} \equiv a^{b} \cdot a^{c}$ $a^(b+c)-=a^b*a^c$

$e^{1 + ln x} = e^{1} \cdot e^{ln x} = x e$ $e^(1+lnx)=e^1*e^lnx=xe$

$\frac{d}{d x} (x e) = e$ $d/dx(xe)=e$

$therefored/dx(e^(1+lnx))=e$

How do you find the derivative of e^(1+lnx)e1+lnxe^(1+lnx)?