Derivative Help

[vivistat6]

Okay so what wohld the answer would look like then? I only see one t so where would the x(t) and y(t) go?
dF/dt=(18x^2)*(-16t^3)+(-40y^4)(6)
Would it be (18(4t^4)^2) *(-16t^3)+(-40(6t))(6)?

 

[BigBeachBanana]

Would it be (18(4t^4)^2) *(-16t^3)+(-40(6t))(6)?

That is the answer, although you can simply further.

I only see one t so where would the x(t) and y(t) go?

You've successfully found the rate of change of F as t change. Not sure what you mean by "where would the x(t) and y(t) go?"?

 

[vivistat6]

That is the answer, although you can simply further.

You've successfully found the rate of change of F as t change. Not sure what you mean by "where would the x(t) and y(t) go?"?

Yay!!! Thank you SO much for all your help! If only I had your brain ?

 

[nasi112]

Okay so what wohld the answer would look like then? I only see one t so where would the x(t) and y(t) go?
dF/dt=(18x^2)*(-16t^3)+(-40y^4)(6)
Would it be (18(4t^4)^2) *(-16t^3)+(-40(6t)^4)(6)?

Remove the red negative sign and place power of \(\displaystyle 4\), then your answer will be correct.

And if you simplified, you would get same as my answer. Where is my chocolate bar?

 

[Steven G]

Correct. The goal is to put everything in terms of t, hence the meaning of the notation [imath]\frac{dF}{\red{dt}}[/imath], i.e. how is F changed as t changes?

But it's x(t)...and y(t) so I would want to plug in those equations for t?

No, you have it backwards. You want to replace x with whatever x(t) equals.​

Btw, I am not sure why my post is in bold-I did not have any option.​