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I have the following problem, which requires the chain rule and can be a little tedious. I'm trying to figure out how to do this in Mathematica, but I can't get a solution. I know the answer, which is 20.085. Here's the problem:

enter image description here

And here's my code:

Clear[x, y, f, t]

x[t_] := Sin[t];

y[t_] := Cos[t]

z[t_] := 2 t^3 + 3

f[t_] := x[t] E^(y[t] z[t])

f'[t]

And Mathematica outputs the following:

E^((3 + 2 t^3) Cos[t]) Cos[t] + 
 E^((3 + 2 t^3) Cos[t]) Sin[t] (6 t^2 Cos[t] - (3 + 2 t^3) Sin[t])

Any thought on how I can structure some code that can be used to solve other of these parameterized functions that would ordinarily require the chain rule?

Thanks so much.

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  • $\begingroup$ (the following)/.t->0 will do the job $\endgroup$ – wuyudi Dec 17 '20 at 7:41
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w = x*E^(y*z);
x = Sin[t];
y = Cos[t];
z = 2 t^3 + 3;
D[w, t] /. t -> 0
% // N

E^3

20.0855

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  • $\begingroup$ Many thanks. One last question. On your penultimate line, where t->0, if you have multiple parameters, like s and t, how do you write a restriction like t =2 and s=3. Do I need braces or brackets? $\endgroup$ – Abcderia Dec 17 '20 at 6:02
  • $\begingroup$ @user27847 such as g[s,t]/.{t->2,s->3} $\endgroup$ – cvgmt Dec 17 '20 at 6:15
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Clear[x, y, f, t]
x[t_] = Sin[t];
y[t_] = Cos[t]
z[t_] = 2 t^3 + 3
f[t_] = x[t] E^(y[t] z[t])
f'[t]
% /. t -> 0
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