text: ## Calculus (3rd Edition)
  
  $$y=4x+1.$$
  We have $$y'=4e^{4x}\Longrightarrow y'(0)=4.$$ Then the slope of the tangent line at $x=0$ is $m= 4$. Hence the equation of the tangent line is given by $$y=mx+c=4x+c.$$ Since $y(0)=1$, then $c=1$, hence the tangent line at $x=0$ is given by $$y=4x+1.$$
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