Differentiating within a for loop
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So i have the following equation, i have to differentiate 3 times using a for loop:
let y =ae^bx sin(cx)
Ive followed a booklet guide i was given and the outputs i have look wrong, theres 2 attemps (images) below, does anyone know where im going wrong
for both images the following has been Initialised
Attempt 1
a = 5, b = 2, c = 4, d = 3
syms x
for y=1:1:3, diff (a*exp(b*x))*sin(c*x),end
ans =
10*sin(4*x)*exp(2*x)
ans =
10*sin(4*x)*exp(2*x)
ans =
10*sin(4*x)*exp(2*x)
Attempt 2
for y=1:1:3, diff a*exp(b*x)*sin(c*x),end
ans =
Columns 1 through 14
-55 59 19 -8 -72 58 -56 78 -79 1 73 -10 5 -70
Columns 15 through 18
59 -57 78 -79
ans =
Columns 1 through 14
-55 59 19 -8 -72 58 -56 78 -79 1 73 -10 5 -70
Columns 15 through 18
59 -57 78 -79
ans =
Columns 1 through 14
-55 59 19 -8 -72 58 -56 78 -79 1 73 -10 5 -70
Columns 15 through 18
59 -57 78 -79
4 Comments
Rik
on 25 Mar 2020
In response to the flag by Ryan ("I would like to delete this question please"):
Why do you want to delete your question? If you weren't allowed to ask for help on your homework you shouldn't have posted it in the first place. In case you decide to edit away all content: don't bother, I made a capture of this page on the Wayback Machine, so there will be a permanent record we can easily restore your edits from.
Answers (1)
James Tursa
on 22 Mar 2020
Edited: James Tursa
on 22 Mar 2020
I assume by "three times" what was meant was diffferentiate the results iteratively three times, not differentiate the original three times ... what would be the point in that? E.g.,
df{1} = a * exp(b*x) * sin(c*x);
df{2} = diff(df{1});
df{3} = diff(df{2});
df{4} = diff(df{3});
Then put the last three lines in a loop.
4 Comments
James Tursa
on 23 Mar 2020
What error message did you get? Did you include all of the symbolic stuff for x and the a, b, c up front in your code?
You shouldn't use df for both the index variable name and at the same time use it for your derivatives. That's trying to use the same name for two different things, right?
A loop would look like this, where k is the index variable name
for k=2:n
df{k} = diff(df{k-1});
end
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