How can I get a derivated function df/dt?

Question

0 个投票

Hello,

I'm trying to calculate derivate functions like z(t)=cos(x(t))+y(t)^2 and I expect answers like z'(t)=-x'(t)*sin(x(t))+2*y(t)*y'(t).

I probably need to use MuPAD but I don't know the commands and what I've found on 'help' isn't exactrly what I want.

Thanks for your answers!

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Answer 1

Star Strider 2015-8-7

0 个投票

In R2015a, I get this result:

syms x(t) y(t) z(t)
z(t)=cos(x(t))+y(t)^2;
dzdt = diff(z, t)
dzdt(t) =
2*y(t)*diff(y(t), t) - sin(x(t))*diff(x(t), t)

Star Strider 2015-8-7

It depends on whether it is an additive or multiplicative constant, and where it is in the expression.

Example 1 (multiplicative):

syms x(t) y(t) z(t) c
z(t) = c * cos(x(t)) + y(t)^2;
dzdt = diff(z, t)
dzdt(t) =
2*y(t)*diff(y(t), t) - c*sin(x(t))*diff(x(t), t)

Example 2 (additive):

syms x(t) y(t) z(t) c
z(t) = c + cos(x(t)) + y(t)^2;
dzdt = diff(z, t)
dzdt(t) =
2*y(t)*diff(y(t), t) - sin(x(t))*diff(x(t), t)

As expected, it disappears if an additive constant.

Example 3 (multiplicative inside expression):

syms x(t) y(t) z(t) c
z(t) = cos(c*x(t)) + y(t)^2;
dzdt = diff(z, t)
dzdt(t) =
2*y(t)*diff(y(t), t) - c*sin(c*x(t))*diff(x(t), t)

Example 4 (additive inside expression):

syms x(t) y(t) z(t) c
z(t) = cos(c + x(t)) + y(t)^2;
dzdt = diff(z, t)
dzdt(t) =
2*y(t)*diff(y(t), t) - sin(c + x(t))*diff(x(t), t)

One of these should work for you.