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kumpel [21]
3 years ago
8

Find the equation of a tangent line at the point on a curve f(x) when x=ef(x)=xln(x)+x

Mathematics
1 answer:
alex41 [277]3 years ago
7 0
\bf f(x)=xln(x)+x\implies \cfrac{dy}{dx}=\stackrel{product~rule}{\left[1\cdot ln(x)+x\cdot \cfrac{1}{x}\right]}+1
\\\\\\
\left. \cfrac{dy}{dx}=ln(x)+2 \right|_{x=e} \implies 1+2\implies 3
\\\\\\
\textit{when x=e, what is \underline{y}?}\qquad f(e)=eln(e)+e\implies f(e)=2e
\\\\\\
(e~~,~~2e)\qquad\qquad \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-2e=3(x-e)
\\\\\\
y-2e=3x-3e\implies y=3x-e
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