Correction check on this please ?

[Mathmasteriw]

so this what i get??
Any good?

 

[Mathmasteriw]

..

 

[Harry_the_cat]

..

Yes! Since \(\displaystyle t^0=1\) then it can be simplified to \(\displaystyle 6t^2 +4\)

 

[Dr.Peterson]

so this what i get??
Any good?

The answer is right, but it's wrong to say [MATH]3(2t)^{2-1}[/MATH]; the 2 shouldn't be inside the parentheses. What you mean is [MATH]3(2t^{3-1})[/MATH] or [MATH]3\cdot2(t^{3-1})[/MATH]

 

[Harry_the_cat]

The answer is right, but it's wrong to say [MATH]3(2t)^{2-1}[/MATH]; the 2 shouldn't be inside the parentheses. What you mean is [MATH]3(2t^{3-1})[/MATH] or [MATH]3\cdot2(t^{3-1})[/MATH]

Yes you're correct Dr P. I missed that.

 

[Steven G]

Same comment for (4t)^1-1.

(4t)^1-1 = (4t)⁰ = 1 not 4

You should have written 4t^1-1 = 4t⁰ = 4*1 = 4

 

[Mathmasteriw]

Great stuff guys thanks to you all for your help once again!!

 

[Deleted member 4993]

Great stuff guys thanks to you all for your help once again!!

Don't forget the original problem:

d/dt [sin(2t^3 + 4t - 2)] = ?

 

[Mathmasteriw]

How dose this look now?
Thanks all for your help

 

[Deleted member 4993]

How dose this look now?
Thanks all for your help

Second line from bottom. (an equation)

What did you write on the left-hand-side of that equation?

 

[Mathmasteriw]

Second line from bottom. (an equation)

What did you write on the left-hand-side of that equation?

Seems to be dv/du

 

[Mathmasteriw]

Second line from bottom. (an equation)

What did you write on the left-hand-side of that equation?

Ahh now I see, should be dv/dt ?

 

[Dr.Peterson]

Ahh now I see, should be dv/dt ?

Not quite. Look carefully, and think before writing. What are you differentiating there? Is it v? or ...

 

[Mathmasteriw]

ahh of course, should be du/dt!