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How to optimize the tangent equation of a curve at a certain point as follows：

How to draw the images of the original function, derivative function and tangent equation into the same coordinate system?

ClearAll["`*"]
f[x_] := Log[1 + x] + x E^-x
g[x_] := D[f[x], x] // FullSimplify
point = {0, f[0]}
point[[1]]
k = D[f[x], x] /. x -> point[[1]]
y - point[[2]] == k (x - point[[1]])
Plot[{f[x], g[x],%}, {x, -1, 2}]

Update 1:

f[x_] = Log[1 + x] + x E^-x;
point = {0, f[0]}
k = D[f[x], x] /. x -> point[[1]]
y - point[[2]] == k (x - point[[1]])
Plot[{f[x], f'[x]}, {x, -1, 4}, 
 Epilog -> {Red, 
   InfiniteLine[{x, f[x]}, D[{x, f[x]}, x]] /. x -> point[[1]]}, 
 PlotLabels -> "Expressions", AspectRatio -> 1]

Update 2:

Clear[f, expr];
f[x_] = Log[1 + x] + x E^-x;
point = {2, f[2]}
k = D[f[x], x] /. x -> point[[1]]
y - point[[2]] == k (x - point[[1]])
expr = {f[x], f'[x], f[x0] + f'[x0] (x - x0) /. x0 -> point[[1]]} // 
   Evaluate;
Plot[expr, {x, -1, 7}, AspectRatio -> 1, 
 PlotLabels -> Thread[{"f(x)", "f'(x)", "y"} == expr], 
 ImageSize -> Full]

Update 3

Clear[f, expr];
f[x_] = Log[1 + x] + x E^-x; point = {0, f[0]} 
k = D[f[x], x] /. x -> point[[1]]
 y - point[[2]] == k (x - point[[1]]) 
expr = {f[x], f'[x], f[x0] + f'[x0] (x - x0) /. x0 -> point[[1]]} // 
  Evaluate; 
Plot[expr, {x, -1, 7}, 
 Epilog -> {Red, PointSize[0.02], Point[point], 
   Text[point, point, {-2, 1}]}, AspectRatio -> 1, 
 PlotLabels -> Thread[{"f(x)", "f'(x)", "y"} == expr], 
 ImageSize -> 500]

Update 4:

Clear[f, expr];
f[x_] = Log[x] + E^(x - 1);
point = {1, f[1]} 
k = D[f[x], x] /. x -> point[[1]]
 y - point[[2]] == k (x - point[[1]]) 
expr = {f[x], f'[x], 
   f[x0] + f'[x0] (x - x0) /. x0 -> point[[1]] // FullSimplify} // 
  Evaluate; Plot[expr, {x, -1, 7}, 
 Epilog -> {Red, PointSize[0.02], Point[point], 
   Text[point, point, {-2, 1}]}, AspectRatio -> 1, 
 PlotLabels -> Thread[{"f(x)", "f'(x)", "y"} == expr], 
 ImageSize -> 500]
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  • 1
    $\begingroup$ What exactly do you mean by optimizing? Also, your choice of tags is befuddling $\endgroup$
    – bmf
    Commented Feb 24, 2023 at 10:39
  • $\begingroup$ What about Normal[Series[f[x], {x, x0, 1}]] ? $\endgroup$
    – Ulrich Neumann
    Commented Feb 24, 2023 at 10:41
  • $\begingroup$ How to draw the images of the original function, derivative function and tangent equation into the same coordinate system? $\endgroup$
    – csn899
    Commented Feb 24, 2023 at 10:42
  • 1
    $\begingroup$ f[x],f'[x]need one argument x but tangent equation needs two arguments x,x0! That's why it's necessary to choose x0 before plotting $\endgroup$
    – Ulrich Neumann
    Commented Feb 24, 2023 at 10:44

2 Answers 2

Reset to default
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Edit

Clear[f,expr];
f[x_] = Log[1 + x] + x E^-x;
expr = {f[x], f'[x], f[x0] + f'[x0] (x - x0) /. x0 -> 0} // Evaluate;
Plot[expr, {x, -1, 4}, AspectRatio -> Automatic, 
 PlotLabels -> Thread[{"f(x)", "f'(x)", "y"} == expr]]

enter image description here

Original

Clear[f];
f[x_] = Log[1 + x] + x E^-x;
Plot[{f[x], D[f[x], x]} // Evaluate, {x, -1, 4}, 
 Epilog -> {Red, InfiniteLine[{x, f[x]}, D[{x, f[x]}, x]] /. x -> 0}, 
 AspectRatio -> Automatic, 
 PlotLabels -> {Style["f(x)", 14, Blue], Style["f'(x)", 14, Blue]}]

enter image description here

Improve this answer
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8
  • $\begingroup$ (+1) but it should be f[x_] = Log[1 + x] + x E^-x; Plot[Evaluate[{f[x], D[f[x], x]}], {x, -1, 4}, Epilog -> {Red, InfiniteLine[{x, f[x]}, D[{x, f[x]}, x]] /. x -> 0}] In the OP it is stated the function, its derivative and the tangent line :-) $\endgroup$
    – bmf
    Commented Feb 24, 2023 at 10:56
  • $\begingroup$ PlotLabels -> "Expressions"Plus, why can't the complete function analytic expression be displayed next to each curve in the image? How to display tangent equations next to tangents $\endgroup$
    – csn899
    Commented Feb 24, 2023 at 11:02
  • $\begingroup$ Can I display the complete analytical formula of the function next to the curve? And f (x)=......, $\endgroup$
    – csn899
    Commented Feb 24, 2023 at 11:10
  • $\begingroup$ The tangent equation should be y==2x $\endgroup$
    – csn899
    Commented Feb 24, 2023 at 11:25
  • 1
    $\begingroup$ @csn899 Clear[f, expr]; f[x_] = Log[1 + x] + x E^-x; point = {0, f[0]}; k = D[f[x], x] /. x -> point[[1]] y - point[[2]] == k (x - point[[1]]); expr = {f[x], f'[x], f[x0] + f'[x0] (x - x0) /. x0 -> point[[1]]} // Evaluate; Plot[expr, {x, -1, 7}, Epilog -> {Red, PointSize[0.02], Point[point], Text[point, point, {-1, 1}]}, AspectRatio -> Automatic, PlotLabels -> Thread[{"f(x)", "f'(x)", "y"} == expr], ImageSize -> 400] $\endgroup$
    – cvgmt
    Commented Feb 25, 2023 at 0:18
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Maybe I am missing something, but I don't see what.

Anyway, with a minimal amendment of the code in the OP we have:

f[x_] := Log[1 + x] + x E^-x
g[x_] := D[f[x], x] // FullSimplify
point = {0, f[0]}
point[[1]]
k = D[f[x], x] /. x -> point[[1]]
p0 = ContourPlot[
   y - point[[2]] == k (x - point[[1]]), {x, 0, 4}, {y, 0, 2}];
p1 = Plot[Evaluate[{f[x], g[x]}], {x, 0, 4}, 
   PlotStyle -> {Red, Green}];
Show[p0, p1]
Show[p1, p0]

plot

p2

Improve this answer
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