## 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 $$y=4x+1.$$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}